Quiz

Hint

Using properties of logs we get:
\(
\begin{aligned}
& \log _5 75-\log _5 3=\log _5 \frac{75}{3}=\log _5 25 \\
& \log _5 25=\log _5 5^2=2 \log _5 5=2 \cdot 1=2
\end{aligned}
\)

Hint

Recall the log rule:
\(
\log (a)+\log (b)=\log (a b)
\)
In this particular case, \(a=4\) and \(b=12\). Thus, our answer is \(\log (4 \cdot 12)=\log (48)\).

Hint

Since the bases of the logs are the same and the logarithms are added, the arguments can be multiplied together. We then simplify the right side of the equation:
\(
\begin{aligned}
& \log _6(n-3)+\log _6(n+2)=\log _3(3) \\
& \log _6(n-3)(n+2)=1
\end{aligned}
\)
The logarithm can be converted to exponential form:
\(
\begin{aligned}
& (n-3)(n+2)=6^1 \\
& n^2-n-6=6 \\
& n^2-n-12=0
\end{aligned}
\)
Factor the equation:
\(
\begin{aligned}
& (n-4)(n+3)=0 \\
& n=4
\end{aligned}
\)
Although there are two solutions to the equation, logarithms cannot be negative. Therefore, the only real solution is 4.

Hint

The rule for the addition of logarithms is as follows:
\(
\log (a b)=\log (a)+\log (b) .
\)
As an application of this, \(\log (5 x)=\log (5)+\log (x)\).

Hint

The logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator.
\(
\log _A\left(\frac{B}{C}\right)=\log _A B-\log _A C
\)
If we encounter two logarithms with the same base, we can likely combine them. In this case, we can use the reverse of the above identity.
\(
\begin{aligned}
& \log _A B-\log _A C=\log _A\left(\frac{B}{C}\right) \\
& A=4, B=7, C=5 \\
& \log _4 7-\log _4 5=\log _4\left(\frac{7}{5}\right)
\end{aligned}
\)

Hint

Using the logarithm rules, exponents within logarithms can be removed and simply multiplied by the remaining logarithm. This expression can be simplified as \(5 \log _2(4) \log _2(4)\)

Hint

\(
\log 2^x=6
\)
Use the power reducing theorem:
\(
\log a^x=b \rightarrow x \log a=b
\)
\(a=2\) and \(b=6\)
\(x \log 2=6\)
\(
\begin{aligned}
& x=\frac{6}{\log 2} \\
& x=19.93
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 3+\log _7 x=8-4 \log _7 x \\
& 5 \log _7 x=5 \\
& \log _7 x=1 \\
& x=7^1=7
\end{aligned}
\)

Hint

\(
\begin{aligned}
5^{\frac{\log _3 x}{1+\log _3 x}}=5^{-1} & \\
\frac{\log _3 x}{1+\log _3 x} & =-1 \\
2 \log _3 \mathrm{x} & =-1 \\
\log _3 \mathrm{x} & =-\frac{1}{2} \\
\mathrm{x} & =3^{-0.5}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \log _3\left(5+4 \cdot \log _2(x-1)\right)=2 \\
& \log _3\left(5+4 \cdot \log _2(x-1)\right)=\log _3 9 \\
& 5+4 \cdot \log _2(x-1)=9 \\
& 4 \cdot \log _2(x-1)=4 \\
& \log _2(x-1)=1 \\
& x-1=2^1 \\
& x=3
\end{aligned}
\)

Hint

\(
\begin{gathered}
\log (\mathrm{x}+5)-\log (\mathrm{x}-1)=\log 10-\log 2 \\
\log \frac{x+5}{x-1}=\log \frac{10}{2} \\
\frac{x+5}{x-1}=5 \\
\mathrm{x}+5=5 \mathrm{x}-5 \\
4 \mathrm{x}=10 \\
\mathrm{x}=5/2
\end{gathered}
\)

Hint

\(
\begin{aligned}
& \log (x+2)+\log (x-7)=2 \cdot \log (x-4) \\
& \log (x+2)(x-7)=\log (x-4)^2 \\
& (x+2)(x-7)=(x-4)^2 \\
& x^2-5 x-14=x^2-8 x+16 \\
& 3 x=30 \\
& x=10
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 2 \log 3 x^2+3 \log 4 x^3=4 \log 2 x^2+4 \log 6 x \\
& \log 9 x^4+\log 64 x^9=\log 16 x^8+\log 1296 x^4 \\
& \log \left(576 x^{13}\right)=\log \left(20736 x^{12}\right) \\
& 576 x^{13}=20736 x^{12} \\
& x=20736 x^{12} / 576 x^{12} \\
& x=36
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \log \sqrt{x+4}-\log \sqrt{x-4}=\log 12-\log 4  \\
& \log \frac{\sqrt{x+4}}{\sqrt{x-4}}=\log \frac{12}{4} \\
& \frac{\sqrt{x+4}}{\sqrt{x-4}}=3 \\
& \frac{x+4}{x-4}=9 \\
& x+4=9(x-4) \\
& x+4=9 x-36 \\
& 8 x=40 \\
& x=5 
\end{aligned}
\)

Hint

Using the product rule:
\(
\log _2(x \cdot(x-3))=2
\)
Changing the logarithm form according to the logarithm definition:
\(
x \cdot(x-3)=2^2
\)
Or
\(
x^2-3 x-4=0
\)
Solving the quadratic equation:
\(
x_{1,2}=[3 \pm \sqrt{ }(9+16)] / 2=[3 \pm 5] / 2=4,-1
\)
Since the logarithm is not defined for negative numbers, the answer is:
\(
x=4
\)

Hint

Using the quotient rule:
\(
\log _3((x+2) / x)=2
\)
Changing the logarithm form according to the logarithm definition:
\(
(x+2) / x=3^2
\)
Or
\(
x+2=9 x
\)
Or
\(
8 x=2
\)
Or
\(
x=0.25
\)

Hint

\(
\log _3(x)+\log _3(x-2)=\log _3(x+10)
\)
Apply Product Rule from Log Rules
\(
\log _3[(x)(x-2)]=\log _3(x+10)
\)
Simplify: \((x)(x-2)=x^2-2 x\)
\(
\log _3\left(x^2-2 x\right)=\log _3(x+10)
\)
Drop the logs, set the arguments (stuff inside the parenthesis) equal to each other
\(
x^2-2 x=x+10
\)
Solve the quadratic equation using the factoring method. But you need to move everything on one side while forcing the opposite side equal to 0 .
\(
\begin{array}{r}
x^2-3 x-10=0 \\
(x-5)(x+2)=0
\end{array}
\)
Set each factor equal to zero, then solve for \(x\).
\(x-5=0\) implies that \(x=5\)

Hint

\(
\log _5(x+2)-\log _5(x)=\log _5(2 x-1)-\log _5(3 x-12)
\)
The difference of logs is telling us to use the Quotient Rule. Convert the subtraction operation outside into a division operation inside the parenthesis. Do it to both sides of the equations.
\(
\log _5\left(\frac{x+2}{x}\right)=\log _5\left(\frac{2 x-1}{3 x-12}\right)
\)
I think we are ready to set each argument equal to each other since we can reduce the problem to have a single log expression on each side of the equation.
\(
\log _5\left(\frac{x+2}{x}\right)=\log _5\left(\frac{2 x-1}{3 x-12}\right)
\)
Drop the logs, and set the arguments (stuff inside the parenthesis) equal to each other. Note that this is a Rational Equation. One way to solve it is to get its Cross Product.
\(
\frac{x+2}{x}=\frac{2 x-1}{3 x-12}
\)
It looks like this after getting its Cross Product.
\(
(x+2)(3 x-12)=(x)(2 x-1)
\)
Simplify both sides by the Distributive Property. At this point, we realize that it is just a Quadratic Equation. No big deal then. Move everything to one side, which forces one side of the equation to be equal to zero.
\(
3 x^2-6 x-24=2 x^2-x
\)
This is easily factorable. Now set each factor to zero and solve for \(x\).
\(
\begin{array}{r}
x^2-5 x-24=0 \\
(x+3)(x-8)=0
\end{array}
\)
So, these are our possible answers.
\(
x=-3 \text { or } x=8
\)
-3 is not possible.

Hint

Keep the log expression on the left, and move all the constants on the right side.
\(
\log _3(\sqrt{-7 x+1})-7=-4
\)
Simplify.
\(
\log _3(\sqrt{-7 x+1})=3
\)
I think we’re ready to transform this log equation into the exponential equation.
\(
\log _3(\sqrt{-7 x+1})=3
\)
The expression inside the parenthesis stays in its current location while the constant 3 becomes the exponent of the log base 3 .
\(
\sqrt{-7 x+1}=3^3
\)
Simplify the right side since \(3^3=27\). What we have here is a simple Radical Equation.

\(
\sqrt{-7 x+1}=27
\)
To get rid of the radical symbol on the left side, square both sides of the equation.
\(
(\sqrt{-7 x+1})^2=(27)^2
\)
After squaring both sides, it looks like we have a linear equation. Just solve it as usual.
\(
\begin{aligned}
-7 x+1 & =729 \\
-7 x & =728 \\
\frac{-7 x}{-7} & =\frac{728}{-7} \\
x & =-104
\end{aligned}
\)
Check your potential answer back into the original equation.
After doing so, you should be convinced that indeed \(x=-104\) is a valid solution.

Hint

First, I’ll expand the square on the right-hand side to be the explicit product of two logs:
\(
\begin{aligned}
& \log _2\left(x^2\right)=\left[\log _2(x)\right]^2 \\
& \log _2\left(x^2\right)=\left[\log _2(x)\right]\left[\log _2(x)\right]
\end{aligned}
\)
Then I’ll apply the log rule to move the “squared” from inside the log on the left-hand side of the equation, taking it out in front of that log as a multiplier:
\(
2 \cdot \log _2(x)=\left[\log _2(x)\right]\left[\log _2(x)\right]
\)
Then I’ll move that term from the left-hand side of the equation to the right-hand side:
\(
0=\left[\log _2(x)\right]\left[\log _2(x)\right]-2 \cdot \log _2(x)
\)
This equation may look bad, but take a close look. It’s nothing more than a factoring exercise at this point. So l’ll factor, and then I’ll solve the factors by using The Relationship:
\(
\begin{aligned}
& 0=\left[\log _2(x)\right]\left[\log _2(x)-2\right] \\
& \log _2(x)=0 \text { or } \log _2(x)-2=0 \\
& 2^0=x \text { or } \log _2(x)=2 \\
& 1=x \text { or } 2^2=x \\
& 1=x \text { or } 4=x
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \ln \left(e^x\right)=\ln \left(e^3\right)+\ln \left(e^5\right) \\
& \ln \left(e^x\right)=\ln \left[\left(e^3\right)\left(e^5\right)\right] \\
& \ln \left(e^x\right)=\ln \left(e^{3+5}\right) \\
& \ln \left(e^x\right)=\ln \left(e^8\right)
\end{aligned}
\)
Comparing the arguments, I get:
\(
\begin{aligned}
& e^x=e^8 \\
& x=8
\end{aligned}
\)

Hint

Use log rule: \(a=\log _b b^a\), then: \(1=\ln \left(e^1\right)=\ln (e) \rightarrow \ln (2 x-1)=\ln (e)\)
When the logs have the same base: \(\log _b f(x)=\log _b g(x)\), then: \(f(x)=g(x)\)
Then: \(\ln (2 x-1)=\ln (e)\), then: \(2 x-1=e \rightarrow x=\frac{e+1}{2}\)

Hint

Since \(3^x\left(2^{2 x}\right)=3^x\left(2^2\right)^x=(3 \times 4)^x=12^x\) the equation becomes
\(
\begin{aligned}
& 12^x=7\left(5^x\right) \\
& \left(\frac{12}{5}\right)^x=7 \\
& \log \left(\frac{12}{5}\right)^x=\log 7 \\
& x \log \left(\frac{12}{5}\right)=\log 7 \\
& x=\frac{\log 7}{\log 12 / 5}=2.27
\end{aligned}
\)

Hint

Recall the property that says if \(\log _b x=\log _b y\) then \(x=y\). Since each logarithm is on opposite sides of the equal sign and each has the same base, 4 in this case, we can use this property to just set the arguments of each equal. Doing this gives,
\(
x^2-2 x=5 x-12
\)
Now all we need to do is solve the equation from Step 1 and that is a quadratic equation that we know how to solve. Here is the solution work.
\(
\begin{aligned}
x^2-2 x & =5 x-12 \\
x^2-7 x+12 & =0 \\
(x-3)(x-4) & =0 \quad \rightarrow \quad x=3, \quad x=4
\end{aligned}
\)

Hint

Recall the property that says if \(\log _b x=\log _b y\) then \(x=y\). That doesn’t appear to have any use here since there are three logarithms in the equation. However, recall that we can combine a sum of logarithms (provide the coefficient of each is a one of course…) as follows,
\(
\ln (x(x+3))=\ln (20-5 x)
\)
We now have only two logarithms and each logarithm is on opposite sides of the equal sign and each has the same base, \(\mathbf{e}\) in this case. Therefore, we can use this property to just set the arguments of each equal. Doing this gives,
\(
x(x+3)=20-5 x
\)
Now, all we need to do is solve the equation from Step 1 and that is a quadratic equation that we know how to solve. Here is the solution work.
\(
\begin{aligned}
x(x+3) & =20-5 x \\
x^2+3 x & =20-5 x \\
x^2+8 x-20 & =0 \\
(x+10)(x-2) & =0 \quad \rightarrow \quad x=-10, \quad x=2
\end{aligned}
\)

x=2 is the possible answer

Hint

\(
\text { Let } x=(0.4)^{-\log_{2.5}}\left\{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots\right\}
\)
\(
=\left(\frac{4}{10}\right)^{-\log_{2.5}}\left\{\frac{\frac{1}{3}}{1-\frac{1}{3}}\right\}=\left(\frac{2}{5}\right)^{-\log_{2.5}\left(\frac{1}{2}\right)}=\left(\frac{5}{2}\right)^{{\log _{5 / 2}}\left(\frac{1}{2}\right)}=\frac{1}{2}
\)

Hint

\(
\begin{aligned}
3 \cdot x^{\log _5 2}+2^{\log _5 x} & =64 \\
3 \cdot 2^{\log _5 x}+2^{\log _5 x} & =64 a^{\log _b x}=x^{\log _b a}, b \neq 1, a, b, x \text { are positive numbers. } \\
4 \cdot 2^{\log _5 x} & =64 \\
2^{\log _5 x} & =4^2=2^4 \\
\log _5 x & =4 \\
x & =5^4=625
\end{aligned}
\)

Hint

\(
a^{\sqrt{\left(\log _a b\right)}}=a^{\sqrt{\log _a b} \times \sqrt{\log _a b} \times \sqrt{\log _b a}}
\)
\(
=a^{\log _a b \cdot \sqrt{(\log_b a)}}
\)
\(
=b^{\sqrt{\log _b a}} \quad a^{\log _a x}=x, a>0, a \neq 1, x>0
\)
\(
\text { Hence, } a^{\sqrt{\left(\log _a b\right)}}-b^{\sqrt{\log_b a)}}=0
\)

Hint

\(
\begin{aligned}
& a =\log _{24}(12) \quad b =\log _{36}(24) \quad c =\log _{48}(36) \\
& 1+ abc =1+\log _{24}(12) \log _{36}(24) \log _{48}(36) \\
& \Rightarrow 1+ abc =1+\frac{\log 12}{\log 24} \times \frac{\log 24}{\log 36} \times \frac{\log _{36}}{\log 48} \\
& \Rightarrow 1+ abc =1+\frac{\log 12}{\log 48}=\frac{\log 48+\log 12}{\log 48} \\
& \Rightarrow 1+ abc =\frac{\log (48 \times 12)}{\log 48}=\frac{\log (24)^2}{\log 48}=2 \frac{\log 24}{\log 48} \\
& \Rightarrow 1+ abc =2 \frac{\log 24}{\log 36} \times \frac{\log 36}{\log 48} \\
& \Rightarrow 1+ abc =2 \log _{36}(24) \log _{48}(36) \\
& \Rightarrow 1+ abc =2 bc
\end{aligned}
\)

Hint

We know that \(\log _m m^n=n \log _m m\) and \(\log _m m=1\)
\(
\therefore \log _m(m)^n=n \times \log _m m=n \times 1=n
\)
\(\log _m(m)^n=n \ldots \text { (i) }\)
\(
\log _4\left[\log _2\left(\log _2\left(\log _3 81\right)\right)\right]
\)
\(
=\log _4\left[\log _2\left(\log _2\left(\log _3 3^4\right)\right)\right]\left(\because 81=3^4\right)
\)
\(
=\log _4\left[\log _2\left(\left(\log _2 4\right)\right)[\text { According to equal (i) }]\right.
\)
\(
\begin{aligned}
& =\log _4\left(\log _2\left(\log _2 2^2\right)\right)\left(\because 4=2^2\right) \\
& =\log _4\left(\log _2 2\right)[\text { According to equal (i) }] \\
& =\log _4(1)\left(\because \log _m m=1\right)=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \quad f(n)=\prod_{i=2}^{n-1} \frac{\log (i+1)}{\log (i)}=\frac{\log (n)}{\log (2)}=\log _2 n \\
& \therefore \quad f\left(2^k\right)=k \\
& \text { Then, } \sum_{k=1}^{100} f\left(2^k\right)=\sum_{k=1}^{100} k=\frac{100 \cdot(100+1)}{2}=5050
\end{aligned}
\)

Hint

\(
\log _3 27 \cdot \log _x 7=\log _{27} x \cdot \log _7 3 \dots(i)
\)
Eq. (i) valid for \(x>0, x \neq 1\)
On solving Eq. (i),
\(
\log _3\left(3^3\right) \cdot \log _x 7=\frac{1}{3} \log _3 x \cdot \log _7 3
\)
\(
\begin{aligned}
9 \cdot \log _x 7 & =\log _7 x \\
9 & =\left(\log _7 x\right)^2 \\
\log _7 x & = \pm 3 \\
x & =7^3 \text { or } x=7^{-3}
\end{aligned}
\)
Then, the least value of \(x\) is \(\frac{1}{7^3}\) i.e., \(7^{-3}\).

Hint

\(
\log _5 120+(x-3)-2 \log _5\left(1-5^{x-3}\right)=-\log _5\left(0.2-5^{x-4}\right)
\)
\(
\begin{aligned}
\log _5(5 \times 24) & +(x-3) \\
& =\log _5\left(1-5^{x-3}\right)^2-\log _5\left(\frac{1-5^{x-3}}{5}\right)
\end{aligned}
\)
\(
\begin{aligned}
1+\log _5 24+(x-3) & =\log _5\left\{5 \cdot\left(1-5^{x-3}\right)\right\} \\
1+\log _5\left(24 \cdot 5^{x-3}\right) & =1+\log _5\left(1-5^{x-3}\right) \\
24 \cdot 5^{x-3} & =1-5^{x-3} \\
25 \cdot 5^{x-3} & =1 \\
5^{x-1} & =5^0 \\
x-1 & =0 \Rightarrow x=1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 49^A+5^B=? \\
& A=1-\log _7 2 \\
& A=\log _7 7-\log _7 2 \\
& A=\log _7 \frac{7}{2} \Rightarrow 7^A=\frac{7}{2} \Rightarrow 49^A=\frac{49}{4} \\
&
\end{aligned}
\)
\(
\text { and } B=-\log _5 4=\log _5\left(\frac{1}{4}\right) \Rightarrow 5^B=\frac{1}{4}
\)
\(
\therefore \quad 49^A+5^B=\frac{49}{4}+\frac{1}{4}=\frac{50}{4}=12.5
\)

Hint

\(
\left(\log _{16} x\right)^2-\log _{16} x+\log _{16} \lambda=0 \dots(1)
\)
\(
\text { Eq. (i) defined for } x>0, \lambda>0\left(\log _{16} x-\frac{1}{2}\right)^2-\frac{1}{4}+\log _{16} \lambda=0
\)
For exactly one solution,
\(
\log _{16} x-\frac{1}{2}=0
\)
\(
\begin{aligned}
\therefore & & -\frac{1}{4}+\log _{16} \lambda & =0 \Rightarrow \log _{16} \lambda=\frac{1}{4} \\
& \text { or } & \lambda & =(16)^{1 / 4}=2
\end{aligned}
\)

Hint

Given, \(y=\log _2 \log _6\left(2^{\sqrt{2 x+1}}+4\right) \dots(i)\)
From Eq. (i) to be defined,
\(
2 x+1>0 \Rightarrow x>-\frac{1}{2} \dots(2)
\)
We find value of \(x\) for which \(y=1\)
\(
\therefore \quad 1=\log _2 \log _6\left(2^{\sqrt{2 x+1}}+4\right)
\)
\(
\begin{aligned}
\log _6\left(2^{\sqrt{2 x+1}}+4\right) & =2 \\
2^{\sqrt{2 x+1}}+4 & =36 \\
2^{\sqrt{2 x+1}} & =32=2^5 \Rightarrow \sqrt{2 x+1}=5 \\
2 x+1 & =25 \Rightarrow x=12
\end{aligned}
\)
So, required point is \((12,1)\).

Hint

Given, equation \(2 \log _x a+\log _{a x} a+3 \log _{a^2 x} a=0 \dots(i)\)
\(
\Rightarrow \quad \frac{2}{\log _a x}+\frac{1}{1+\log _a x}+\frac{3}{2+\log _a x}=0 \dots(ii)
\)
Let \(\log _a x=t\)
Then, Eq. (ii),
\(
\frac{2}{t}+\frac{1}{1+t}+\frac{3}{2+t}=0 \Rightarrow 6 t^2+11 t+4=0
\)
\(\Rightarrow \quad t=-\frac{4}{3}\) or \(-\frac{1}{2}\)
So, \(x=a^{-4 / 3}\) or \(x=a^{-1 / 2}\)
Two value of \(x\) possible for which Eq. (i) is defined and satisfy.

Hint

\(
\begin{aligned}
a & =\log _3 \log _3 2 \\
3^a & =\log _3 2 \\
3^{-a} & =\log _2 3
\end{aligned}
\)
\(
\begin{array}{lc}
\text { Now, } & 1<2^{-k+3^{-a}}<2^1 \\
\Rightarrow & 2^0<2^{-k+3^{-a}}<2^1 \\
\therefore & 0<-k+3^{-a}<1 \\
\Rightarrow & 0<-k+\log _2 3<1 \\
\Rightarrow & 0>k-\log _2 3>-1 \\
\Rightarrow & \log _2 3-1<k<\log _2 3 \\
\therefore & k=1
\end{array}
\)

Hint

\(
\because(2 x)^{\ln 2}=(3 y)^{\ln 3}
\)
Taking log with base \(e\) on both sides, then
\(
\begin{aligned}
\ln 2(\ln 2+\ln x) & =\ln 3(\ln 3+\ln y) \dots(i) \\
\text { and } \quad 3^{\ln x} & =2^{\ln y}
\end{aligned}
\)
Taking log with base \(e\) on both sides, then
\(
\ln x \cdot \ln 3=\ln y \cdot \ln 2 \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\ln 2(\ln 2+\ln x)=\ln 3\left(\ln 3+\frac{\ln x \cdot \ln 3}{\ln 2}\right)
\)
\(
\begin{aligned}
\ln x\left(\frac{(\ln 3)^2}{\ln 2}-\ln 2\right) & =-\left((\ln 3)^2-(\ln 2)^2\right) \\
\ln x=-\ln 2 & =\ln \left(\frac{1}{2}\right) \\
x & =\frac{1}{2} \\
x_0 & =\frac{1}{2}
\end{aligned}
\)

Hint

\(
\text { Let } S=\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \cdots \infty}}}
\)
\(
\begin{aligned}
S & =\frac{1}{3 \sqrt{2}} \sqrt{4-S} \\
(3 \sqrt{2} S)^2 & =4-S \\
18 S^2+S-4 & =0 \\
(9 S-4)(2 S+1) & =0 \\
9 S-4 & =0 [\because 2 S+1 \neq 0]
\end{aligned}
\)
\(
\begin{aligned}
S & =\frac{4}{9}=\left(\frac{3}{2}\right)^{-2} \\
\log _{3 / 2} S & =-2 \Rightarrow 6+\log _{3 / 2} S=6-2=4
\end{aligned}
\)
Hence,
\(
6+\log _{3 / 2}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \cdots}}}\right)=4
\)

Hint

choice (a,b,c) are the answers.

\(
3^x=4^{x-1}
\)
taking \(\log\) on both sides with base a
\(
x \log _a 3=(x-1) \log _a 4 \Rightarrow x=\frac{\log _a 4}{\log _a 4-\log _a 3}
\)
If \(a=3: x=\frac{2 \log _3 2}{2 \log _1 2-1}\)
If \(a =2: x =\frac{2}{2-\log _2 3}\)
If \(a =4: x=\frac{1}{1-\log _4 3}\)

Hint

\(
\begin{aligned}
(\sqrt{7})^{\frac{1}{\log _4 7}} & =(\sqrt{7})^{\log _7 4} \\
& =(7)^{\frac{1}{2} \log _7 4} \\
& =(4)^{\frac{1}{2} \log _7 7} \\
& =(4)^{\frac{1}{2}}=2
\end{aligned}
\)
\(
\left(\left(\log _2 9\right)^2\right)^\frac{1}{\log _2\left(\log _2 9\right)}
\)
We know \(\log _a x=\frac{1}{\log _x a}\)
\(
=\left(\left(\log _2 9\right)^2\right)^{\log _{\left(\log _2 9\right)} 2}
\)
\(
=\left(\log _2 9\right)^{2 \log _{\left(\log _2 9\right)} 2}
\)
We know \(a^{\log _a x}=x\) and applying this we get
\(
=\left(\log _2 9\right)^{\log _{\left(\log _2 9\right)} 2^2}
\)
\(
=2^2=4
\)
The value of \(\left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2 \log _2 9}} \times(\sqrt{7})^{\log _4 7}\) is=4 x 2 =8

Hint

\(
\begin{aligned}
& \log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1 \\
& \Rightarrow \log _{\cos x} \cot x-4 \log _{\sin x} \cot x=1 \\
& \Rightarrow 1-\log _{\cos x} \sin x-4-4 \log _{\sin x} \cos x=1
\end{aligned}
\)
Let \(\log _{\cos x} \sin x=t\)
\(
\begin{aligned}
& t+\frac{4}{t}=4 \\
& \Rightarrow t=2 \\
& \sin x=\cos ^2 x \\
& \Rightarrow \sin x=1-\sin ^2 x \\
& \Rightarrow \sin ^2 x+\sin x^{-1}=0 \\
& \Rightarrow \sin x=\frac{-1 \pm \sqrt{5}}{2} \\
& \text { as } x \in\left(0, \frac{\pi}{2}\right) \\
& \sin x=\frac{\sqrt{5}-1}{2} \\
& x=\sin ^{-1}\left(\frac{-1+\sqrt{5}}{2}\right) \\
& \Rightarrow \alpha=-1, \beta=5 \\
& \alpha+\beta=4
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \log _{(x+1)}\left(2 x^2+7 x+5\right)+\log _{(2 x+5)}(x+1)^2-4=0 \\
& \log _{(x+1)}(2 x+5)(x+1)+2 \log _{(2 x+5)}(x+1)=4 \\
& \log _{(x+1)}(2 x+5)+1+2 \log _{(2 x+5)}(x+1)=4 \\
& \text { Put } \log _{(x+1)}(2 x+5)=t
\end{aligned}
\)
\(
\begin{aligned}
& t+\frac{2}{t}=3 \Rightarrow t^2-3 t+2=0 \\
& t =1,2 \\
& \log _{(x+1)}(2 x+5)=1 \& \log _{(x+1)}(2 x+5)=2 \\
& x+1=2 x+5 \& 2 x+5=(x+1)^2 \\
& x=-4 \text { (rejected) } \\
& x^2=4 \Rightarrow x=2,-2 \text { (rejected) }
\end{aligned}
\)
So, \(x=2\)
No. of solution \(=1\)

Hint

\(
\begin{aligned}
& \log _4(x-1)=\log _2(x-3) \\
& \Rightarrow \frac{1}{2} \log _2(x-1)=\log _2(x-3) \\
& \Rightarrow \log _2(x-1)^{1 / 2}=\log _2(x-3)
\end{aligned}
\)
\(
\begin{aligned}
& (x-1)^{1 / 2}=x-3 \\
& x-1=x^2+9-6 x \\
& x^2-7 x+10=0 \\
& (x-2)(x-5)=0 \\
& x=2,5
\end{aligned}
\)
But \(x \neq 2\) because it is not satisfying the domain of given equation i.e. \(\log _2(x-3) \rightarrow\) its domain \(x>\) 3
finally \(x\) is 5
\(\therefore\) No. of solutions \(=1\).

Hint

\(
\begin{aligned}
\frac{1}{\log _2 n} & +\frac{1}{\log _3 n}+\frac{1}{\log _4 n}+\ldots . . .+\frac{1}{\log _{1983} n} \\
& =\log _n 2+\log _n 3+\log _n 4+\ldots . .+\log _n 1983
\end{aligned}
\)
\(
=\log _n(2.3 .4 \ldots .1983)=\log _n(1983!)=\log _n n=1
\)

Hint

\(
\begin{aligned}
& x=(\sqrt{2}+1)^{1 / 3}-(\sqrt{2}-1)^{1 / 3} \\
& x^3=(\sqrt{2}+1)-(\sqrt{2}-1)-3(\sqrt{2}+1)^{1 / 3}(\sqrt{2}-1)^{1 / 3}
\end{aligned}
\)
\(
[\sqrt[3]{(\sqrt{2}+1)}-\sqrt[3]{\sqrt{2}-1}]
\)
\(
x^3=2-3(2-1)^{1 / 3} x \Rightarrow x^3+3 x=2
\)

Hint

\(
\log _{0.3}(x-1)<\log _{0.09}(x-1)
\)
\(
\begin{aligned}
& 1<\frac{\log _{0.09}(x-1)}{\log _{0.3}(x-1)} \Rightarrow 1<\log _{0.3}(0.09) \\
& \Rightarrow 1<\log _{0.3}(0.3)^2 \Rightarrow 1<2
\end{aligned}
\)
Which of true therefore it is true for every positive value of 2 .
\(
\therefore x \in(1, \infty) \text {. }
\)

Hint

We have \(2 \log 10 x-\log x(0.01), x>1\)
\(
\begin{aligned}
& =2 \log _{10} x-\log _x(10)^{-2} \\
& =2\left(\log _{10} x+\log _x(10)\right) \\
& 2\left(\log _{10} x+\frac{1}{\log _{10} x}\right) \dots(1)
\end{aligned}
\)
Now, use A. M. \(\geq\) G. M. that is arithmetic mean is greater than or equal to geometric mean of two numbers.
\(
\begin{aligned}
& \Rightarrow \frac{\left(\log _{10} x+\frac{1}{\log _{10} x}\right)}{2} \geq \sqrt{\log _{10} x \times \frac{1}{\log _{10} x}} \\
& \frac{\left(\log _{10} x+\frac{1}{\log _{10} x}\right)}{2} \geq 1
\end{aligned}
\)
\(
\begin{aligned}
& \left(\log _{10} x+\frac{1}{\log _{10} x}\right) \geq 2 \\
& 2\left(\log _{10} x+\frac{1}{\log _{10} x}\right) \geq 4
\end{aligned}
\)
So, from 1 minimum value of \(2 \log _{10} x-\log _x(0.01)\) is 4.

Hint

\(
\begin{array}{r}
y=\log_{10} x \\
x=10^y \dots(1)
\end{array}
\)
Reflected
\(
y=10^x \dots(2)
\)
Comparing eqn(1) and eqn(2), we get \(y=x\)

Hint

\(
\text { Applying, A.M. } \geq \text { G.M., }
\)
\(
\text { we get, } \frac{\log _a x+\log _x a}{2} \geq \sqrt{\log _a x \cdot \log _x a}
\)
We know that \(\log _a x=\frac{1}{\log _x a}\)
\(
\frac{\log _\alpha x+\log _x a}{2} \geq 1
\)
\(
\left(\log _x a+\frac{1}{\log _x a}\right) \geq 2
\)
\(
\text { Thus, the minimum value of }\left(\log _{ a } x+\log _{ x } a\right) \text { is } 2 \text {. }
\)

Hint

The given equation is
\(
\begin{array}{ll}
& \log _7\left[\log _5 \sqrt{x+5}+\sqrt{x}\right]=0 \\
\Rightarrow & \log _5(\sqrt{x+5}+\sqrt{x})=7^0=1 \\
\Rightarrow & \sqrt{x+5}+\sqrt{x}=5^1=5 \\
\Rightarrow & \sqrt{x+5}=5-\sqrt{x} \\
\Rightarrow & x+5=25-10 \sqrt{x}+x \\
\Rightarrow & 10 \sqrt{x}=20 \\
\Rightarrow & \sqrt{x}=2 \\
\Rightarrow & x=4
\end{array}
\)
Hence, the solution is \(x=4\).

Hint

The given equation is
\(
\begin{aligned}
& \log _{(2 x+3)}\left(6 x^2+23 x+21\right)=4-\log _{(3 x+7)}\left(4 x^2+12 x+9\right) \\
& \Rightarrow \quad \log _{(2 x+3)}(2 x+3)(3 x+7)=4-\log _{(3 x+7)}(2 x+3)^2 \\
& \Rightarrow \quad 1+\log _{(2 x+3)}(3 x+7)=4-2 \log _{(3 x+7)}(2 x+3)
\end{aligned}
\)
\(
\begin{aligned}
& \log _{(2 x+3)}(3 x+7)=3-\frac{2}{\log _{(2 x+3)}(3 x+7)} \\
& \quad \Rightarrow \quad y=3-\frac{2}{y}, \text { where } y=\log _{(2 x+3)}(3 x+7) \\
& y^2-3 y+2=0 \\
& (y-1)(y-2)=0 \\
& y=1,2
\end{aligned}
\)
\(
\text { When } y=1
\)
\(
\begin{aligned}
& \log _{(2 x+3)}(3 x+7)=1 \\
\Rightarrow \quad & 3 x+7=2 x+3 \\
\Rightarrow \quad & x=-4
\end{aligned}
\)
When \(y=2\),
\(
\begin{aligned}
& \log _{(2 x+3)}(3 x+7)=1 \\
& 3 x+7=(2 x+3)^2 \\
& \quad=4 x^2+12 x+9 \\
& 4 x^2+9 x+2=0 \\
& 4 x^2+8 x+x+2=0 \\
& 4 x(x+2)+1(x+2)=0 \\
& (x+2)(4 x+1)=0 \\
& x=-2,-\frac{1}{4}
\end{aligned}
\)
As \(x>-\frac{3}{2}\), so \(x=-\frac{1}{4}\)
Hence, the solution is \(x=-\frac{1}{4}\).

Hint

We have,
\(
x^{(3 / 4)\left(\log _2 x\right)^2+\log _2 x-\frac{5}{4}}=\sqrt{2}
\)
\(
\begin{aligned}
& \left((3 / 4)\left(\log _2 x\right)^2+\log _2 x-\frac{5}{4}\right) \log x=\log (\sqrt{2}) \\
& \left((3 / 4) b^2+b-\frac{5}{4}\right) b=\frac{1}{2}, \text { where } b=\log _2 x \\
& 3 b^3+4 b^2-5 b-2=0 \\
& 3 b^3-3 b^2+7 b^2-7 b+2 b-2=0 \\
& 3 b^2(b-1)+7 b^2-7 b+2 b-2=0 \\
& (b-1)\left(\left(3 b^2+7 b+2\right)=0\right. \\
& (b-1)\left(3 b^2+6 b+b+2\right)=0 \\
& (b-1)[3 b(b+2)+1(b+2)]=0 \\
& (b-1)(b+2)(3 b+1)=0 \\
& b=1,-2,-\frac{1}{3} \\
& \log _2 x=1,-2,-\frac{1}{3} \\
& x=2,2^{-2}, 2^{-\frac{1}{3}}
\end{aligned}
\)
Thus, the equation has exactly three real solutions of which exactly one is irrational.

Hint

\(
\begin{aligned}
& \text { Given } \log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \in AP \\
& \Rightarrow \quad 2 \log _3\left(2^x-5\right)=\log _3 2+\log _3\left(2^x-\frac{7}{2}\right) \\
& \Rightarrow \quad \log _3\left(2^x-5\right)^2=\log _3 2 \cdot\left(2^x-\frac{7}{2}\right) \\
& \Rightarrow \quad\left(2^x-5\right)^2=2 \cdot\left(2^x-\frac{7}{2}\right) \\
& \Rightarrow \quad\left(2^x\right)^2-10 \times 2^x+25=2 \times 2^x-7 \\
& \Rightarrow \quad\left(2^x\right)^2-12 \times 2^x+32=0 \\
& \Rightarrow \quad(a)^2-12 \times a+32=0 \\
& \Rightarrow \quad(a-8)(a-4)=0 \\
& \Rightarrow \quad a=8,4
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad 2 x=8,4=2^3, 2^2 \\
& \Rightarrow \quad x=3,2
\end{aligned}
\)
Since \(x=2\) does not satisfy the logarithmic expression,
\(
\therefore \quad x=3
\)
Hence, the solution is \(x=3\).

Hint

Let \(\log _2 7=\frac{p}{q}\), where \(p, q \in N\) and \(\operatorname{HCF}(p, q)=1\)
\(
\begin{array}{ll}
\Rightarrow & 2^{\frac{p}{q}}=7 \\
\Rightarrow & 2^p=7^q
\end{array}
\)
This is not possible for any \(p, q \in N\) and 7 and 2 are prime.
Thus, \(\log _2 7\) is an irrational number.

Hint

As per trigonometric properties:
\(\tan \theta=\cot (90-\theta)\) and \(\tan \theta \cdot \cot \theta=1\)
thus, \(\tan 89^{\circ}=\cot \left(90-89^{\circ}\right)=\cot 1^{\circ}\)
similarly, \(\tan 88^{\circ}=\cot 2^{\circ}, \tan 87^{\circ}=\cot 3^{\circ}\) ans so on…
also from Logarithm properties:
\(
\begin{aligned}
& \log m +\log n =\log ( m \times n ) \\
& \text { thus, } \log \tan 1^{\circ}+\log \tan 89^{\circ}=\log \left(\tan 1^{\circ} \times \tan 89^{\circ}\right) \\
& =\log \left(\tan 1^{\circ} \times \cot 1^{\circ}\right) \\
& =\log 1 \\
& =0
\end{aligned}
\)
similarly, \(\log \tan 2^{\circ}+\log \tan 88^{\circ}=\log \left(\tan 1^{\circ} \times \cot 1^{\circ}\right)=\log 1=0\)
thus, \(\log \tan 1^{\circ}+\log \tan 2^{\circ}+\ldots \ldots .+\log \tan 89^{\circ}=0\).

Hint

\(
\begin{aligned}
& \log _3 4 \cdot \log _4 5 \cdot \log _5 6 \cdot \log _6 7 \cdot \log _7 8 \cdot \log _8 9 \\
& =\log _3 4 \cdot \log _4 5 \cdot \log _5 6 \cdot \log _6 7 \cdot \log _7 8 \cdot \log _8 9 \\
& =\log _3 9=\log _3\left(3^2\right)=2 \log _3(3)=2
\end{aligned}
\)

Alternate:

\(
\log _b a=\frac{\log _b a}{\log b}
\)
\(
\frac{\log _4}{\log _3} \cdot \frac{\log _5}{\log _4} \cdot \frac{\log _6}{\log _5} \cdot \frac{\log _7}{\log _6} \cdot \frac{\log _8}{\log 7} \cdot \frac{\lg 9}{\log _8}
\)
\(
\frac{\log 9}{\log 3}=\frac{\log 3^2}{\log 3}=\frac{2 \log 3}{\log 3}=(2)
\)

Hint

\(
\begin{aligned}
& \log _3 4 \cdot \log _4 5 \cdot \log _5 10 \cdot \log _{10} 32 \\
& =\log _2(32) \\
& =\log _2\left(2^5\right) \\
& =5
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \log _k x \cdot \log _5 k=\log _x 5 \\
& \frac{\log x}{\log k} \times \frac{\log k}{\log 5}=\frac{\log 5}{\log x} \\
& (\log x)^2=(\log 5)^2 \\
& (\log x)^2-(\log 5)^2=0 \\
& \{(\log x)+(\log 5)\}\{(\log x)-(\log 5)\}=0 \\
& \text { Either }(\log x)+(\log 5))=0 \text { or }(\log x)-(\log 5)=0
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & \log (5 x)=0 \text { or }\left(\log \left(\frac{x}{5}\right)\right)=0 \\
\Rightarrow & \log (5 x)=\log 1 \text { or }\left(\log \left(\frac{x}{5}\right)\right)=\log 1 \\
\Rightarrow & 5 x=1 \text { or }\left(\frac{x}{5}\right)=1 \\
\Rightarrow & x=\frac{1}{5} \text { or } 5
\end{array}
\)
Therefore, the solutions are \(\left\{5, \frac{1}{5}\right\}\).

Hint

\(
\begin{aligned}
A & =\log _2 \log _2 \log _4 256 \\
& =\log _2 \log _2 \log _4 256 \\
& =\log _2 \log _2 \log _4\left(4^4\right) \\
& =\log _2\left\{\left(\log _2 4\right)\left(\log _4 4\right)\right\} \\
& =\log _2\left(\log _2 4\right) \\
& =\log _2\left(\log _2 2^2\right) \\
& =\left(\log _2 2\right)\left(\log _2 2\right)=1
\end{aligned}
\)
and
\(
\begin{aligned}
B & =2 \log _{\sqrt{2}} 2=2 \log _{2^{1 / 2}}(2) \\
& =2 \times 2\left(\log _2 2\right)=4
\end{aligned}
\)
Therefore, the value of
\(
A+B+10=1+4+10=15
\)

Hint

Given \(\log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right)\) are in AP.
\(
\begin{aligned}
& \Rightarrow \quad 2 \log _3\left(2^x-5\right)=\log _3 2+\log _3\left(2^x-\frac{7}{2}\right) \\
& \Rightarrow \quad \log _3\left(2^x-5\right)^2=\log _3 2 \cdot\left(2^x-\frac{7}{2}\right) \\
& \Rightarrow \quad\left(2^x-5\right)^2=2 \times 2^x-7 \\
& \Rightarrow \quad\left(2^x\right)^2-12 \times 2^x+32=0 \\
& \Rightarrow \quad a^2-12 a+32=0 \text { where } a=2^x \\
& \Rightarrow \quad(a-4)(a-8)=0 \\
& \Rightarrow \quad a=4,8
\end{aligned}
\)
When \(a=4 \Rightarrow 2^x=4=2^2 \Rightarrow x=2\) when \(a=8 \Rightarrow 2^x=8=2^3 \Rightarrow x=3\) But \(x=2\) does not satisfy the terms. Hence, the solution of \(x\) is 3.

Hint

Let \(\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{a \log a}{a(b-c)}=\frac{b \log b}{b(c-a)}=\frac{c \log c}{c(a-b)}=k \\
& \Rightarrow \quad \frac{\log a^a}{a(b-c)}=\frac{\log b^b}{b(c-a)}=\frac{\log c^c}{c(a-b)}=k \\
& \Rightarrow \quad \log \left(a^a\right)+\log \left(b^b\right)+\log \left(c^c\right) \\
& \quad=k(a b-b c+b c-b a+c a-c b) \\
& \Rightarrow \quad=0 \\
& \Rightarrow \quad \log \left(a^a b^b c^c\right)=0 \\
& \Rightarrow \quad\left(a^a b^b c^c\right)=e^0=1
\end{aligned}
\)

Hint

Given
\(
\begin{array}{ll}
& x=\log _a b c \\
\Rightarrow & a_x=b c \\
\Rightarrow & a \times a^x=a b c \\
\Rightarrow & a^{x+1}=a b c
\end{array}
\)
\(
\begin{aligned}
& \Rightarrow \quad x+1=\log _a(a b c) \\
& \Rightarrow \quad \frac{1}{x+1}=\frac{1}{\log _a(a b c)} \\
& \text { Similarly, } \frac{1}{y+1}=\frac{1}{\log _b(a b c)} \text {, } \\
& \text { and } \quad \frac{1}{z+1}=\frac{1}{\log _c(a b c)} \\
& \text { Thus, } \quad \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1} \\
& =\frac{1}{\log _a(a b c)}+\frac{1}{\log _b(a b c)}+\frac{1}{\log _c(a b c)} \\
& =\log _{a b c}(a)+\log _{a b c}(b)+\log _{a b c}(c) \\
& =\log _{a b c}(a b c) \\
& =1 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
a b & =\log _{12} 18 \times \log _{24} 54 \\
& =\frac{\log 18}{\log 12} \times \frac{\log 54}{\log 24} \\
& =\frac{\log \left(2 \times 3^2\right)}{\log \left(3 \times 2^2\right)} \times \frac{\log \left(2 \times 3^3\right)}{\log \left(3 \times 2^3\right)} \\
& =\frac{\log 2+2 \log 3}{\log 3+2 \log 2} \times \frac{\log 2+3 \log 3}{\log 3+3 \log 2} \\
& =\frac{1+2 \log _2 3}{\log _2 3+2} \times \frac{1+3 \log _2 3}{\log _2 3+3} \\
& =\frac{1+2 x}{x+2} \times \frac{1+3 x}{x+3} \text { where } x=\log _2 3 \\
& =\frac{1+5 x+6 x^2}{x^2+5 x+6}
\end{aligned}
\)
\(
\begin{aligned}
(a-b) & =\log _{12} 18-\log _{24} 54 \\
& =\frac{\log \left(2 \times 3^2\right)}{\log \left(3 \times 2^2\right)}-\frac{\log \left(2 \times 3^3\right)}{\log \left(3 \times 2^3\right)} \\
& =\frac{\log 2+2 \log 3}{\log 3+2 \log 2}-\frac{\log 2+3 \log 3}{\log 3+3 \log 2} \\
& =\frac{1+2 \log _2 3}{\log _2 3+2}-\frac{1+3 \log _2 3}{\log _2 3+3} \\
& =\frac{1+2 x}{x+2}-\frac{1+3 x}{x+3} \text { where } x=\log _2 3 \\
& =\frac{\left(1-x^2\right)}{(2+x)(3+x)}
\end{aligned}
\)
Hence, the value of
\(
\begin{aligned}
5(a-b)+a b & =\frac{5\left(1-x^2\right)}{(2+x)(3+x)}+\frac{\left(1+5 x+6 x^2\right)}{(2+x)(3+x)} \\
& =\frac{5\left(1-x^2\right)+\left(1+5 x+6 x^2\right)}{(2+x)(3+x)} \\
& =\frac{x^2+5 x+6}{(2+x)(3+x)} \\
& =\frac{(x+2)(x+3)}{(2+x)(3+x)} \\
& =1
\end{aligned}
\)

Hint

Given
\(
\begin{aligned}
& \log _2 x+\log _2 y \geq 6 \\
& \Rightarrow \quad \log _2(x y) \geq 6 \\
& \Rightarrow \quad x y \geq 2^6=64
\end{aligned}
\)
As we know that, \(\frac{x+y}{2} \geq \sqrt{x y}\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{x+y}{2} \geq \sqrt{64}=8 \\
& \Rightarrow \quad x y \geq 2.8=16
\end{aligned}
\)
Hence, the least value of \(x+y\) is 16 .

Hint

Given,
\(
\begin{aligned}
& x^{18}=y^{21}=z^{28} \\
\Rightarrow \quad & \log \left(x^{18}\right)=\log \left(y^{21}\right)=\log \left(z^{28}\right) \\
\Rightarrow \quad & 18 \log x=21 \log y=28 \log z=k \text { (say) }
\end{aligned}
\)
Now, \(3 \log _y x=3 \cdot \frac{\log x}{\log y}=3 \cdot \frac{21}{18}=\frac{7}{2}\)
\(
3 \log _z y=3 \cdot \frac{\log y}{\log z}=3 \cdot \frac{28}{21}=4
\)
and \(7 \log _x z=7 \cdot \frac{\log z}{\log x}=7 \cdot \frac{18}{28}=\frac{9}{2}\)
Thus, \(3,7 / 2,4,9 / 2\) are in AP.

Hint

Given equation is
\(
4^{\log _9 3}+9^{\log _2 4}=10^{\log _x 83}
\)
\(
\begin{array}{ll}
\Rightarrow & 4^{\frac{1}{2} \log _3 3}+9^{2 \log _2 2}=10^{\log _x 83} \\
\Rightarrow & 2+81=10^{\log _x 83} \\
\Rightarrow & 83=83^{\log _x 10} \\
\Rightarrow & \log _x 10=1 \\
\Rightarrow & x=10
\end{array}
\)
Hence, the solution is \(x=10\).

Hint

We have,
\(
\begin{aligned}
a b c & =\log _{24} 12 \times \log _{36} 24 \times \log _{48} 12 \\
& =\log _{48} 36 \times \log _{36} 24 \times \log _{24} 12 \\
& =\log _{48} 12
\end{aligned}
\)
Now,
\(
\begin{aligned}
a b c+1 & =\log _{48} 12+1=\log _{48} 12+\log _{48} 48 \\
& =\log _{48}(12 \times 48)
\end{aligned}
\)
Also, \(b c=\log _{36} 24 \times \log _{48} 36\)
\(
\begin{aligned}
& =\log _{48} 36 \times \log _{36} 24 \\
& =\log _{48} 24
\end{aligned}
\)
\(
\text { Thus, } \begin{aligned}
\left(\frac{a b c+1}{b c}\right) & =\frac{\log _{48}(12 \times 48)}{\log _{48} 24} \\
& =\log _{24}(12 \times 48) \\
& =\log _{24}(24 \times 48) \\
& =\log _{24}\left(24^2\right) \\
& =2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Given } \quad \log _{10}\left(\sin \left(x+\frac{\pi}{4}\right)\right)=\frac{1}{2}\left(\log _{10} 6-1\right) \\
& \Rightarrow \quad 2 \log _{10}\left(\sin \left(x+\frac{\pi}{4}\right)\right)=\left(\log _{10} 6-1\right) \\
& \Rightarrow \quad 2 \log _{10}\left(\frac{1}{\sqrt{2}}(\sin x+\cos x)\right)=\left(\log _{10} 6-1\right) \\
& \Rightarrow \quad 2 \log _{10}\left(\frac{1}{\sqrt{2}}\right)+2 \log _{10}(\sin x+\cos x)
\end{aligned}
\)
\(
=\log _{10} 6-1
\)
\(
\begin{aligned}
& \Rightarrow \quad 2 \log _{10}(\sin x+\cos x)=\log _{10} 6+2 \log _{10} 2-1 \\
& \Rightarrow \quad \log _{10}(\sin x+\cos x)^2=\log _{10}\left(\frac{24}{10}\right) \\
& \Rightarrow \quad(\sin x+\cos x)^2=\left(\frac{24}{10}\right) \\
& \Rightarrow \quad 1+\sin 2 x=\frac{24}{10} \\
& \Rightarrow \quad \sin 2 x=\frac{24}{10}-1=\frac{14}{10} \\
& \Rightarrow \quad \sin x \cdot \cos x=\frac{7}{10} \\
& \Rightarrow \quad \log _{10}(\sin x \cos x)=\log _{10}\left(\frac{7}{10}\right) \\
& \Rightarrow \quad \log _{10}(\sin x)+\log _{10}(\cos x)=\log _{10} 7-1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& a^4 b^5=1 \\
&= \log \left(a^4 b^5\right)=\log (1)=0 \\
=&= 4 \log a+5 \log b=0 \\
&= 4 \log a=-5 \log b \\
&= \frac{\log a}{\log b}=-\frac{5}{4} \\
&= \log _b a=-\frac{5}{4}
\end{aligned}
\)
Now,
\(
\begin{aligned}
\log _a\left(a^5 b^4\right) & =5 \log _a a+4 \log _a b \\
& =5+4 \log _a b \\
& =5+\frac{4}{\log _b a} \\
& =5-\frac{4 \times 4}{5}=\frac{25-16}{5}=\frac{9}{5}
\end{aligned}
\)

Hint

Given
\(
\begin{aligned}
& \log _x\left(\log _{18}(\sqrt{2}+\sqrt{8})\right)=-\frac{1}{2} \\
\Rightarrow & \log _x\left[\log _{18}(\sqrt{2}+2 \sqrt{2})\right]=-\frac{1}{2} \\
\Rightarrow & \log _x\left[\log _{(3 \sqrt{2})^2}(3 \sqrt{2})\right]=-\frac{1}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad \log _x\left(\frac{1}{2}\right)\left[\log _{(3 \sqrt{2})}(3 \sqrt{2})\right]=-\frac{1}{2} \\
& \Rightarrow \quad \log _x\left(\frac{1}{2}\right)=-\frac{1}{2} \\
& \Rightarrow \quad-\log _x 2=-\frac{1}{2} \\
& \Rightarrow \quad \log _x 2=\frac{1}{2} \\
& \Rightarrow \quad \sqrt{x}=2 \\
& \Rightarrow \quad x=4
\end{aligned}
\)

Hint

\(
\log _6 9-\log _9 27+\log ^8 x=\log _{64} x-\log _6 4
\)
\(
\begin{aligned}
& \Rightarrow \quad \log _8 x-\log _{64} x=\log _9 27-\log _6 9-\log _6 4 \\
& \Rightarrow \quad \log _8 x-\frac{1}{2} \log _8 x=\log _{3^2}(3)^3-\log _6 36
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{1}{2} \log _8 x=\frac{3}{2}-\log _6\left(6^2\right) \\
& \Rightarrow \quad \frac{1}{2} \log _8 x=\frac{3}{2}-2=-\frac{1}{2} \\
& \Rightarrow \quad \frac{1}{2} \log _8 x=-\frac{1}{2} \\
& \Rightarrow \quad \log _8 x=4 \\
& \Rightarrow \quad x=\frac{1}{8}
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
12+6 \sqrt{3} & =12+2 \times 3 \times \sqrt{3} \\
& =3^2+(\sqrt{3})^2+2 \times 3 \times \sqrt{3} \\
& =(3+\sqrt{3})^2
\end{aligned}
\)
Similarly, \(12-6 \sqrt{3}=(3-\sqrt{3})^2\)
Now,
\(
\begin{aligned}
x & =\sqrt{12+6 \sqrt{3}}+\sqrt{12-6 \sqrt{3}} \\
& =(3+\sqrt{3})+(3-\sqrt{3}) \\
& =6
\end{aligned}
\)
Thus, \(\log _{36} x=\log _{36} 6\)
\(
=\log _{6^2} 6=\frac{1}{2} \log _6 6=\frac{1}{2}
\)

Hint

We have
\(
\begin{aligned}
& \log _a b \times \log _b c=2 \times 2=4 \\
\Rightarrow \quad & \log _a c=4 \\
\Rightarrow \quad & \frac{\log _3 c}{\log _3 a}=4 \dots(i)
\end{aligned}
\)
Also, \(\log _3 c=3+\log _3 a\)
\(
\begin{aligned}
& \Rightarrow \quad 4 \log _3 a=3+\log _3 a \text { [from (i)] }\\
& \Rightarrow \quad 3 \log _3 a=3 \\
& \Rightarrow \quad \log _3 a=3 \\
& \Rightarrow \quad a=3
\end{aligned}
\)
\(
\begin{aligned}
& \text { Again, } \log _a b=2 \\
& \Rightarrow \quad \log _3 b=2 \\
& \Rightarrow \quad b=3^2=9
\end{aligned}
\)
Further, \(\log _b c=2\)
\(
\begin{aligned}
& \Rightarrow \quad \log _9 c=2 \\
& \Rightarrow \quad c=9^2=81
\end{aligned}
\)
Thus, \((a+b+c)+7=3+9+81+7=100\)

Hint

\(
\begin{aligned}
& \text { Given } 3^{x+2}=45 \\
& \Rightarrow \quad x+2=\log _3(45) \\
& =\log _3(5 \times 9) \\
& =\log _3 5+\log _3 9 \\
& =\log _3 5+2 \\
& \Rightarrow \quad x=\log _3 5 \\
& \Rightarrow \quad x=\frac{\log _{10} 5}{\log _{10} 3} \\
& \Rightarrow \quad x=\frac{\log _{10}\left(\frac{10}{2}\right)}{\log _{10} 3}=\frac{\log _{10} 10-\log _{10} 2}{\log _{10} 3} \\
& \Rightarrow \quad x=\frac{1-a}{b} \\
&
\end{aligned}
\)

Hint

We have,
\(
\begin{aligned}
N & =6 \log _{10} 2+\log _{10} 31 \\
& =\log _{10} 2^6+\log _{10} 31 \\
& =\log _{10}(64 \times 31) \\
& =\log _{10}(1984) \\
& <\log _{10}(1000)=3
\end{aligned}
\)
Also, \(N=\log _{10}(1984)>\log _{10}(10000)=4\)
Thus, the sum of successive integers \(=3+4=7\).

Hint

We have
\(
\begin{aligned}
M & =\log _{\sqrt{2}}^2\left(\frac{1}{4}\right)=\left(\log _{\sqrt{2}}\left(2^{-2}\right)\right)^2 \\
& =\left(-\frac{2}{1 / 2} \log _2 2\right)^2=(-4)^2=16
\end{aligned}
\)
and
\(
\begin{aligned}
N & =\log _{2 \sqrt{2}}^3(8)=\left[\log _{2 \sqrt{2}}(8)\right]^3 \\
& =\left[\log _{2 \sqrt{2}}(2 \sqrt{2})^3\right]^3=\left[3 \log _{2 \sqrt{2}}(2 \sqrt{2})\right]^3 \\
& =(3)^3=27
\end{aligned}
\)
Also,
\(
\begin{aligned}
& P=\log _5\left(\log _3(\sqrt{\sqrt[5]{9}})\right) \\
& =\log _5\left[\log _3\left(9^{1 / 10}\right)\right] \\
& =\log _5\left[\log _3\left(3^{2 / 10}\right)\right] \\
& =\log _5\left(\frac{1}{5}\right)\left[\log _3(3)\right] \\
& =\log _5\left(5^{-1}\right)=-1
\end{aligned}
\)
Thus,
\(
\begin{aligned}
\left(\frac{M}{N}+P+3\right) & =\frac{16}{27}-1+3 \\
& =\frac{16}{27}+2 \\
& =\frac{70}{27}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { We have } \log _{10}(x-2)+\log _{10} y=0 \\
& \Rightarrow \quad \log _{10}[y(x-2)]=\log _{10} 1 \\
& \Rightarrow \quad y(x-2)=1 \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Also, } \sqrt{x}+\sqrt{y-2}=\sqrt{x+y} \\
& \Rightarrow \quad x+y-2+2 \sqrt{x(y-2)}=x+y \\
& \Rightarrow \quad-2+2 \sqrt{x(y-2)}=0 \\
& \Rightarrow \quad-2=-2 \sqrt{x(y-2)} \\
& \Rightarrow \quad x(y-2)=1 \dots(ii)
\end{aligned}
\)
From Eqs (i) and (ii), we get
\(
x=y
\)
Put \(x=y\) in Eq. (ii), we get
\(
x(x-2)=1
\)
\(
\begin{aligned}
& x^2-2 x-1=0 \\
& x^2-2 x+1=2 \\
& (x-1)^2=2 \\
& x-1= \pm \sqrt{2} \\
& x=1 \pm \sqrt{2} \\
& x=1+\sqrt{2}=y
\end{aligned}
\)
Thus, the value of
\(
\begin{aligned}
& x+y-2 \sqrt{2} \\
= & 1+\sqrt{2}+1+\sqrt{2}-2 \sqrt{2} \\
= & 2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \log _{27} a+\log _9 b=\frac{7}{2} \\
\Rightarrow & \log _{3^3} a+\log _{3^2} b=\frac{7}{2} \\
\Rightarrow & \frac{1}{3} \log _3 a+\frac{1}{2} \log _3 b=\frac{7}{2} \\
\Rightarrow & 2 \log _3 a+3 \log _3 b=21 \dots(i)
\end{aligned}
\)
Also, \(\log _{27} b+\log _9 a=\frac{2}{3}\)
\(
\Rightarrow \quad \frac{1}{3} \log _3 b+\frac{1}{2} \log _3 a=\frac{2}{3}
\)
\(
\Rightarrow \quad 2 \log _3 b+3 \log _3 a=4 \dots(ii)
\)
Solving Eqs (i) and (ii), we get
\(
4 \log _3 a-9 \log _3 a=42-12
\)
\(
\begin{aligned}
&= -5 \log _3 a=30 \\
&= \log _3 a=-6 \\
&= a=3^{-6}
\end{aligned}
\)
From Eq. (ii), we get
\(
2 \log _3 b-18=5
\)
\(
\begin{aligned}
& \Rightarrow \quad 2 \log _3 b=2 \\
& \Rightarrow \quad \log _3 b=1 \\
& \Rightarrow \quad b=3^{11}
\end{aligned}
\)
Hence, the value of \(a b=3^{-6} \times 3^{11}=243\).

Hint

Given equation is
\(
\begin{array}{ll}
& \log _{\tan x}\left(2+4 \cos ^2 x\right)=2 \\
\Rightarrow \quad & 2+4 \cos ^2 x=\tan ^2 x \\
\Rightarrow & 2+4 \cos ^2 x=\frac{\sin ^2 x}{\cos ^2 x}
\end{array}
\)
\(
\begin{aligned}
& \Rightarrow \quad 4 \cos ^4 x+2 \cos ^2 x-\sin ^2 x=0 \\
& \Rightarrow \quad 4 \cos ^4 x+3 \cos ^2 x-1=0 \\
& \Rightarrow \quad 4 \cos ^4 x+2 \cos ^2 x-\cos ^2 x-1=0 \\
& \Rightarrow \quad 4 \cos ^2 x\left(\cos ^2 x+1\right)-\left(\cos ^2 x+1\right)=0 \\
& \Rightarrow \quad\left(4 \cos ^2 x-1\right)\left(\cos ^2 x+1\right)=0 \\
& \Rightarrow \quad\left(4 \cos ^2 x-1\right)=0 \quad(\because x \in[0,2 \pi)
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad \cos ^2 x=\left(\frac{1}{2}\right)^2=\cos ^2\left(\frac{\pi}{3}\right) \\
& \Rightarrow \quad x=n \pi \pm \frac{\pi}{3}, n=0,1,2 \\
& \Rightarrow \quad x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}
\end{aligned}
\)
Hence, the number of values of \(x\) is 4 .

Hint

Given,
\(
\begin{aligned}
& \cos (\ln x)=0 \\
\Rightarrow \quad & \ln x=\frac{\pi}{2} \quad \Rightarrow \quad x=e^{\frac{\pi}{2}}
\end{aligned}
\)
Thus, the value of
\(
\begin{aligned}
& \left(\frac{2}{\pi} \times \log (x)+10\right) \\
& =\left(\frac{2}{\pi} \times \log \left(e^{\frac{\pi}{2}}\right)+10\right) \\
& =\left(\frac{2}{\pi} \times \frac{\pi}{2} \log (e)+10\right) \\
& =1+10=11
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Given, } c(a-b)=a(b-c) \\
& \Rightarrow \quad a c-b c=a b-a c \\
& \Rightarrow \quad 2 a c=a b+b c=b(a+c) \\
& \Rightarrow \quad b=\frac{2 a c}{(a+c)}
\end{aligned}
\)
Now,
\(
\begin{aligned}
& \frac{\log (a+c)+\log (a-2 b+c)}{\log (a-c)} \\
& =\frac{\log \{(a+c)(a+c-2 b)\}}{\log (a-c)} \\
& =\frac{\log \left\{(a+c)\left(a+c-\frac{4 a c}{(a+c)}\right)\right\}}{\log (a-c)} \\
& =\frac{\log \left\{(a+c)^2-4 a c\right\}}{\log (a-c)} \\
& =\frac{\log (a-c)^2}{\log (a-c)} \\
& =\frac{2 \log (a-c)}{\log (a-c)}=2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Given } 10^x+10^{-x}=4 \\
& \Rightarrow \quad 10^x+\frac{1}{10^x}=4 \\
& \Rightarrow \quad\left(10^x\right)^2-4\left(10^x\right)+1=0 \\
& \Rightarrow \quad 10^x=\frac{4 \pm \sqrt{16-4}}{2} \\
& \Rightarrow \quad 10^x=\frac{4 \pm \sqrt{12}}{2} \\
& \Rightarrow \quad 10^x=2+\sqrt{3} \\
& \Rightarrow \quad x=\log _{10}(2+\sqrt{3})
\end{aligned}
\)
Thus \(A=2, B=3\)
Hence, the value of \(A+B+3=2+3+3=8\).

Hint

We have
\(
\begin{aligned}
& \log _{10} x+\log x^{1 / 2}+\log _{10} x^{1 / 4}+\ldots=y \\
\Rightarrow \quad & \log _{10} x+\frac{1}{2} \log _{10} x+\frac{1}{4} \log _{10} x+\ldots=y \\
\Rightarrow \quad & y=\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right) \log _{10} x
\end{aligned}
\)
\(
\begin{aligned}
& =\left(\frac{1}{1-\frac{1}{2}}\right) \log _{10} x=2 \log _{10} x \\
& \text { Also, } \frac{1+3+5+\ldots+(2 y-1)}{4+7+10+\ldots+(3 y+1)}=\frac{20}{7 \log _{10} x} \\
& \Rightarrow \frac{\frac{y}{2}(1+2 y-1)}{\frac{y}{2}(4+3 y+1)}=\frac{20}{7\left(\frac{y}{2}\right)} \\
& \Rightarrow \quad \frac{2 y}{(3 y+5)}=\frac{40}{7 y} \\
& \Rightarrow \quad \frac{y}{(3 y+5)}=\frac{20}{7 y} \\
&
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & 7 y^2=60 y+100 \\
\Rightarrow & 7 y^2-60 y-100=0 \\
\Rightarrow & (y-10)(10 y+7)=0 \\
\Rightarrow & y=10,-10 / 7
\end{array}
\)
\(
\begin{aligned}
& \text { When } y=10, \\
& \quad 2 \log _{10} x=10 \\
& \Rightarrow \quad \log _{10} x=5 \\
& \Rightarrow \quad x=10^5 \\
& \text { When } y=-10 / 7, \\
& \quad 2 \log _{10} x=-10 / 7 \\
& \Rightarrow \quad \log _{10} x=-5 / 7 \\
& \Rightarrow \quad x=10^{-5 / 7}
\end{aligned}
\)

Hint

We have
\(
\frac{6}{5} a^{\log _a x \log _{10} a \log _a 5-3 \log _{10}\left(\frac{x}{10}\right)}=9^{\log _{100} x+\log _4 2}
\)
\(
=\frac{6}{5} a^{\frac{\log x}{\log a} \cdot \frac{\log 5}{\log 10}-3\left(\log _{10} x-1\right)}=9^{\log _{100} x+\log _{2^2} 2}
\)
\(
\begin{aligned}
&= \frac{6}{5} a^{\log _a x \log _{10} 5 .-3\left(\log _{10} x-1\right)}=9^{\log _{10^2} x+\log _{2^2} 2} \\
&= \frac{6}{5} a^{\log _a x \log _{10} 5 .-3\left(\log _{10} x-1\right)}=9^{\frac{1}{2}\left(\log _{10} x+1\right)} \\
&= \frac{6}{5} a^{\log _a x \log _{10} 5 .-3\left(\log _{10} x-1\right)}=3^{\left(\log _{10} x+1\right)} \\
&= \frac{2}{5} a^{\log _a x \log _{10} 5} \cdot a^{3-3 \log _{10} x}=3^{\log _{10} x} \\
&= \frac{2}{5} a^{\log _{10} x \log _a 5} \cdot a^{3-3 \log _{10} x}=3^{\log _{10} x} \\
&= \frac{2}{5} a^{b \log _a 5} \cdot a^{3-3 b}=3^b, b=\log _{10} x
\end{aligned}
\)
It is possible only when, \(b=1\)
i.e., \(\log _{10} x=1\)
\(
\Rightarrow \quad x=10^1=10
\)
Hence, the solution is \(x=10\).

Hint

We have
\(
\begin{aligned}
& \log _{100}|x+y|=\frac{1}{2} \\
\Rightarrow & \log _{10^2}|x+y|=\frac{1}{2} \\
\Rightarrow & \frac{1}{2} \log _{10}|x+y|=\frac{1}{2} \\
\Rightarrow & \log _{10}|x+y|=1 \\
\Rightarrow & |x+y|=10 \\
\Rightarrow & x+y= \pm 10 \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Also, } \log _{10}\left(\frac{y}{|x|}\right)=\log _{100} 4 \\
& \Rightarrow \quad \log _{10}\left(\frac{y}{|x|}\right)=\log _{10^2} 2^2 \\
& \Rightarrow \quad \log _{10}\left(\frac{y}{|x|}\right)=\log _{10} 2 \\
& \Rightarrow \quad \frac{y}{|x|}=2
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad|x|=\frac{y}{2} \\
& \Rightarrow \quad x= \pm \frac{y}{2} \dots(ii)
\end{aligned}
\)
From Eqs (i) and (ii), we get
\(
\begin{aligned}
y & = \pm \frac{20}{3}, \pm 20 \\
\text { and } \quad x & = \pm \frac{10}{3}, \mp 10
\end{aligned}
\)
Hence, the solutions are
\(
\left\{x=\frac{10}{3}, y=\frac{20}{3} ; x=10, y=-20\right\} .
\)

Hint

\(
\begin{aligned}
& 2 \log _2 \log _2 x+\log _{1 / 2} \log _2(2 \sqrt{2} x)=1 \\
\Rightarrow \quad & 2 \log _2 \log _2 x-\log _2 \log _2(2 \sqrt{2} x)=1 \\
\Rightarrow \quad & \log _2\left(\log _2 x\right)^2-\log _2\left(\log _2(2 \sqrt{2} x)\right)=1 \\
\Rightarrow \quad & \log _2\left(\frac{\left(\log _2 x\right)^2}{\left(\log _2(2 \sqrt{2} x)\right)}\right)=1 \\
\Rightarrow \quad & \left(\frac{\left(\log _2 x\right)^2}{\left(\log _2(2 \sqrt{2} x)\right)}\right)=2 \\
\Rightarrow \quad & \left(\frac{\left(\log _2 x\right)^2}{\left(\log _2(2 \sqrt{2})+\log _2 x\right)}\right)=2
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad\left(\frac{\left(\log _2 x\right)^2}{\left(\log _2\left(2^{3 / 2}\right)+\log _2 x\right)}\right)=2 \\
& \Rightarrow \quad\left(\frac{\left(\log _2 x\right)^2}{\left(\frac{3}{2}+\log _2 x\right)}\right)=2
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad\left(\frac{a^2}{\frac{3}{2}+a}\right)=2, \text { where } a=\log _2 x \\
& \Rightarrow \quad \frac{2 a^2}{3+2 a}=2 \\
& \Rightarrow \quad 2 a^2-4 a-6=0 \\
& \Rightarrow \quad a^2-2 a-3=0 \\
& \Rightarrow \quad(a-3)(a+1)=0 \\
& \Rightarrow \quad a=-1,3
\end{aligned}
\)
When \(a=-1 \Rightarrow \log _2 x=-1 \Rightarrow x=2^{-1}=\frac{1}{2}\)
When \(a=3 \Rightarrow \log _2 x \Rightarrow x=2^3=8\)
Hence, the solution is \(x=8\).

Hint

\(
\begin{aligned}
& \log _{3 / 4} \log _8\left(x^2+7\right)+\log _{1 / 2} \log _{1 / 4}\left(x^2+7\right)^{-1}=-2 \\
\Rightarrow \quad & \log _{3 / 4} \log _{2^3}\left(x^2+7\right)+\log _{2^{-1}} \log _{2^{-2}}\left(x^2+7\right)^{-1}=-2
\end{aligned}
\)
\(
\Rightarrow \quad \log _{3 / 4}\left(\frac{1}{3}\left(\log _2\left(x^2+7\right)\right)\right)-\log _2\left(\frac{1}{2} \log _2\left(x^2+7\right)\right)=-2
\)
\(
\Rightarrow \quad \log _{3 / 4}\left(\frac{1}{3} y\right)-\log _2\left(\frac{1}{2} y\right)=-2 \text {, }\text { where } y=\left(\log _2\left(x^2+7\right)\right)
\)
\(
\begin{array}{ll}
\Rightarrow & \log _{3 / 4}\left(\frac{y}{3}\right)-\log _2\left(\frac{y}{2}\right)=-2 \\
\Rightarrow & y=4, \text { by trial } \\
\Rightarrow & \log _2\left(x^2+7\right)=4 \\
\Rightarrow & x^2+7=2^4=16 \\
\Rightarrow & x^2=9 \\
\Rightarrow & x= \pm 3
\end{array}
\)
Hence, the solutions are \(\{-3,3\}\).

Hint

We have,
\(
\begin{aligned}
y & =\log _2 x+\log _4 x+\log _8 x+\ldots \\
& =\log _2 x+\frac{1}{2} \log _2 x+\frac{1}{4} \log _2 x+\ldots \\
& =\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right) \log _2 x \\
& =\left(\frac{1}{1-\frac{1}{2}}\right) \log _2 x=2 \log _2 x
\end{aligned}
\)
\(
\begin{aligned}
& \text { Also, } \frac{5+9+13+\ldots+(4 y+1)}{1+3+5+\ldots+(2 y-1)}=4 \log _4 x \\
& \Rightarrow \quad \frac{\frac{y}{2}(2.5+(y-1) 4)}{\frac{y}{2}(2 .+(y-1) 2)}=4 \log _4 x \\
& \Rightarrow \quad \frac{(10+4 y-4)}{(2+2 y-2)}=4 \log _4 x \\
& \Rightarrow \quad \frac{(6+4 y)}{(2 y)}=4 \log _4 x \\
& \Rightarrow \quad \frac{(3+2 y)}{y}=4 \log _{2^2} x=2 \log _2 x=y
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad y^2=2 y+3 \\
& \Rightarrow \quad y^2-2 y-3=0 \\
& \Rightarrow \quad(y-3)(y+1)=0 \\
& \Rightarrow \quad y=-1,3 \\
& \Rightarrow \quad y=3(y=-1 \text { is not possible })
\end{aligned}
\)
\(
\begin{aligned}
& \text { When } y=3, \\
& 2 \log _2 x=3 \\
& \Rightarrow \quad \log _2 x=\frac{3}{2} \\
& \Rightarrow \quad x=2^{3 / 2}
\end{aligned}
\)
Hence, the solutions are \(x=2^{3 / 2}\) and \(y=3\).

Hint

Given,
\(
\begin{array}{ll}
& \log _{225} x+\log _{64} y=4 \\
\Rightarrow \quad & \log _{15^2} x+\log _{8^2} y=4 \\
\Rightarrow \quad & \log _{15} x+\log _8 y=4 \dots(i)
\end{array}
\)
\(
\begin{aligned}
& \text { Also, } \log _x(225)-\log _y(64)=1 \\
& \Rightarrow \quad \log _x(15)-\log _y(8)=\frac{1}{2} \dots(ii)
\end{aligned}
\)
From Eqs (i) and (ii), we get
\(
\log _x(15)-\frac{1}{8-\log _{15}(x)}=\frac{1}{2}
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{1}{\log _{15}(x)}-\frac{1}{8-\log _{15}(x)}=\frac{1}{2} \\
& \Rightarrow \quad \frac{1}{p}-\frac{1}{8-p}=\frac{1}{2} \text { where } p=\log _{15}(x) \\
& \Rightarrow \quad \frac{1}{p}-\frac{1}{8-p}=\frac{1}{2} \\
& \Rightarrow \quad \frac{8-p-p}{p(8-p)}=\frac{1}{2} \\
& \Rightarrow \quad \frac{8-2 p}{p(8-p)}=\frac{1}{2} \\
& \Rightarrow \quad 16-4 p=8 p-p^2 \\
& \Rightarrow \quad p^2-12 p+16=0 \\
& \Rightarrow \quad p=6 \pm 2 \sqrt{5} \\
& \Rightarrow \quad \log _{15} x=6 \pm 2 \sqrt{5} \\
& \Rightarrow \quad x=15^{(6 \pm 2 \sqrt{5})}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { and } & \log _8 y=8-\log _{15} x \\
\Rightarrow & \log _8 y=8-(6 \pm 2 \sqrt{5}) \\
\Rightarrow & \log _8 y=2 \pm 2 \sqrt{5} \\
\Rightarrow & y=8^{2 \pm 2 \sqrt{5}}
\end{array}
\)
Hence, the solutions are
\(
\left(15^{6+2 \sqrt{5}}, 8^{2-2 \sqrt{5}}\right),\left(15^{6-2 \sqrt{5}}, 8^{2+2 \sqrt{5}}\right)
\)
Now,
\(
\begin{aligned}
& \frac{1}{2} \log _{30}(a b c d) \\
& =\frac{1}{2} \log _{30}\left(15^{6+2 \sqrt{5}+6-2 \sqrt{5}} \times 8^{2-2 \sqrt{5}+2+2 \sqrt{5}}\right) \\
& =\frac{1}{2} \log _{30}\left(15^{12} \times 8^4\right) \\
& =\frac{1}{2} \log _{30}\left(15^{12} \times 2^{12}\right) \\
& =\frac{1}{2} \log _{30}(15 \times 2)^{12} \\
& =\frac{12}{2} \log _{30}(30) \\
& =6
\end{aligned}
\)

Hint

We have,
\(
\begin{aligned}
x & =1+\log _a \\
& =\log _a a+\log _a b c=\log _a a b c \\
\Rightarrow \quad \frac{1}{x} & =\frac{1}{\log _a a b c} \\
\Rightarrow \quad \frac{1}{x} & =\log _{a b c} a
\end{aligned}
\)
Similarly, \(\frac{1}{y}=\log _{a b c} b\)
and \(\frac{1}{z}=\log _{a b c} c\)
Now, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\log _{a b c} a+\log _{a b c} b+\log _{a b c} c\)
\(
\begin{aligned}
& =\log _{a b c}(a b c) \\
& =1
\end{aligned}
\)
Hence, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{x y+y z+z x}{x y z}=1 \\
& \Rightarrow \quad \frac{x y z}{x y+y z+z x}=1
\end{aligned}
\)

Hint

We have
\(
\begin{aligned}
& \sum_{r=0}^{n-1} \log _2\left(\frac{r+2}{r+1}\right)=\prod_{r=10}^{99} \log _r(r+1) \\
& \log _2\left(\frac{2}{1}\right)+\log _2\left(\frac{3}{2}\right)+\log _2\left(\frac{4}{3}\right)+\ldots+\log _2\left(\frac{n+1}{n}\right) \\
& =\log _{10}(11) \log _{11}(12) \log _{12}(13) \ldots \log _{99}(100) \\
& =\log _{10}(100)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Thus, } \log _2\left(\frac{2}{1}\right)+\log _2\left(\frac{3}{2}\right)+\log _2\left(\frac{4}{3}\right)+\ldots+\log _2\left(\frac{n+1}{n}\right) \\
& \quad=\log _{10}(100) \\
& \Rightarrow \quad \log _2\left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \ldots \frac{n+1}{n}\right)=2 \log _{10}(10)=2 \\
& \Rightarrow \quad \log _2\left(\frac{n+1}{1}\right)=2 \\
& \Rightarrow \quad n+1=4 \\
& \Rightarrow \quad n=3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 3 \log _x 4+2 \log _{4 x} 4+3 \log _{16 x} 4=0 \\
&= \frac{3}{\log _4 x}+\frac{2}{\log _4(4 x)}+\frac{3}{\log _4(16 x)}=0 \\
&= \frac{3}{\log _4 x}+\frac{2}{1+\log _4 x}+\frac{3}{2+\log _4 x}=0 \\
&= \frac{3}{a}+\frac{2}{1+a}+\frac{3}{2+a}=0 \text { where } a=\log _4 x \\
&= \frac{3}{a}+\frac{3}{2+a}=-\frac{2}{1+a} \\
&= \frac{6+3 a+3 a}{a(2+a)}=-\frac{2}{1+a}
\end{aligned}
\)
\(
\begin{aligned}
&= \frac{6(1+a)}{a(2+a)}=-\frac{2}{1+a} \\
&= 6(1+a)^2+2 a(a+2)=0 \\
&= 6 a^2+12 a+6+2 a^2+4 a=0 \\
&= 8 a^2+16 a+6=0 \\
&= 4 a^2+8 a+3=0 \\
&= 4 a^2+6 a+2 a+3=0 \\
&= 2 a(2 a+3)+1(2 a+3)=0 \\
&= (2 a+3)(2 a+1)=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad a=-\frac{3}{2},-\frac{1}{2} \\
& \Rightarrow \quad \log _4 x=-\frac{3}{2},-\frac{1}{2} \\
& \Rightarrow \quad x=4^{-3 / 2}, 4^{-1 / 2} \\
& \Rightarrow \quad \alpha=\frac{1}{8}, \beta=\frac{1}{2}
\end{aligned}
\)
Hence, the value of
\(
\frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{1}{2}(8+2)=5
\)

Hint

Given equation is
\(
\begin{aligned}
& |x-2|^{\log _2\left(x^3\right) 3 \log _x 4}=(x-2)^3 \\
\Rightarrow \quad & \log _2\left(x^3\right)-3 \log _x 4=3 \\
\Rightarrow \quad & 3 \log _2 x-\frac{3}{\log _4 x}=3
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & \log _2 x-\frac{2}{\log _2 x}=1 \\
\Rightarrow & a-\frac{2}{a}=1 \text { where } a=\log _2 x \\
\Rightarrow & a^2-a-2=0 \\
\Rightarrow & (a-2)(a+1)=0 \\
\Rightarrow & a=2-1 \\
\Rightarrow & \log _2 x=2,-1 \\
\Rightarrow & x=4, \frac{1}{2}
\end{array}
\)
Hence, the value of \((\alpha+2 \beta+3)\) is 8 .

Hint

Given equation is
\(
\begin{aligned}
& 6\left(\log _x 2-\log _4 x\right)+7=0 \\
\Rightarrow & 6\left(\log _x 2-\frac{1}{\log _x 4}\right)+7=0 \\
\Rightarrow & 6\left(\log _x 2-\frac{1}{2 \log _x 2}\right)+7=0 \\
\Rightarrow & 6\left(a-\frac{1}{2 a}\right)+7=0, a=\log _x 2
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & 6\left(2 a^2-1\right)+14 a=0 \\
\Rightarrow & 3\left(2 a^2-1\right)+7 a=0 \\
\Rightarrow & 6 a^2+7 a-3=0 \\
\Rightarrow & 6 a^2+9 a-2 a-3=0 \\
\Rightarrow & 3 a(2 a+3)-(2 a+3)=0 \\
\Rightarrow & (3 a-1)(2 a+3)=0 \\
\Rightarrow & a=\frac{1}{3},-\frac{3}{2} \\
\Rightarrow & \log _x 2=\frac{1}{3},-\frac{3}{2} \\
\Rightarrow & \frac{1}{\log _2 x}=\frac{1}{3},-\frac{3}{2} \\
\Rightarrow & \log _2 x=3,-\frac{2}{3} \\
\Rightarrow & x=2^3, 2^{-2 / 3}
\end{array}
\)
Thus the integral solution of \(x\) is 8 .
Therefore, \(\alpha=8\)
Hence, the value of \(\left(\frac{2 \alpha-1}{5}\right)\)
\(
=\left(\frac{16-1}{5}\right)=3
\)

Hint

Given equation is
\(
\begin{aligned}
& \log \left|\frac{x^2-x-1}{x^2+x-2}\right|=0 \\
\Rightarrow \quad & \left|\frac{x^2-x-1}{x^2+x-2}\right|=1 \\
\Rightarrow \quad & \left|x^2-x-1\right|=\left|x^2+x-2\right| \\
\Rightarrow \quad & x^2-x-1= \pm\left(x^2+x-2\right)
\end{aligned}
\)
Taking positive sign, we get
\(
\begin{aligned}
& \Rightarrow \quad\left(x^2-x-1\right)=\left(x^2+x-2\right) \\
& \Rightarrow \quad 2 x=1 \quad \Rightarrow \quad x=1 / 2
\end{aligned}
\)
Taking negative sign, we get
\(
\begin{aligned}
& \left(x^2-x-1\right)=\left(x^2+x-2\right) \\
\Rightarrow \quad & 2 x^2=3 \Rightarrow x= \pm \sqrt{\frac{3}{2}}
\end{aligned}
\)
Hence, the solutions are
\(
\left\{-\sqrt{\frac{3}{2}}, \frac{1}{2}, \sqrt{\frac{3}{2}}\right\}
\)

Hint

Given equation is
\(
\begin{array}{ll}
& \left|4+\log _{1 / 7} x\right|=2+\left|2+\log _{1 / 7} x\right| \\
\Rightarrow & \left|4+\log _{1 / 7} x\right|=|2|+\left|2+\log _{1 / 7} x\right| \\
\Rightarrow & 2\left(2+\log _{1 / 7} x\right) \geq 0 \\
\Rightarrow & \left(2+\log _{1 / 7} x\right) \geq 0 \\
\Rightarrow & \log _{1 / 7} \geq-2 \\
\Rightarrow & x \leq\left(\frac{1}{7}\right)^{-2} \\
\Rightarrow & x \leq 49
\end{array}
\)
Hence, the value of \(x[latex] is [latex](0,49]\).

Hint

Given equation is
\(
\begin{aligned}
& \log ^2\left(1+\frac{4}{x}\right)+\log ^2\left(1-\frac{4}{x+4}\right)=2 \log ^2\left(\frac{2}{x-1}-1\right) \\
\Rightarrow \quad & \log ^2\left(\frac{x+4}{x}\right)+\log ^2\left(\frac{x}{x+4}\right)=2 \log ^2\left(\frac{2}{x-1}-1\right) \\
\Rightarrow \quad & \log ^2\left(\frac{x+4}{x}\right)+\log ^2\left(\frac{x}{x+4}\right)=2 \log ^2\left(\frac{3-x}{x-1}\right) \\
\Rightarrow \quad & -2 \log \left(\frac{x}{x+4}\right) \log \left(\frac{x+4}{x}\right)=2 \log ^2\left(\frac{3-x}{x-1}\right) \\
\Rightarrow \quad & -\log \left(\frac{x}{x+4}\right) \log \left(\frac{x+4}{x}\right)=\log ^2\left(\frac{3-x}{x-1}\right) \\
\Rightarrow \quad & \log ^2\left(\frac{x}{x+4}\right)=\log ^2\left(\frac{3-x}{x-1}\right) \\
\Rightarrow \quad & \left(\frac{x}{x+4}\right)=\left(\frac{3-x}{x-1}\right) \\
\Rightarrow \quad & x^2-x=3 x+12-x^2-4 x
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & 2 x^2=12 \\
\Rightarrow & x^2=6 \\
\Rightarrow & x= \pm \sqrt{6}
\end{array}
\)
Hence, the solution is \(\{-\sqrt{6}, \sqrt{6}\}\).