Rules and Properties of Logarithm

Product rule

The base remains the same, the sum of the logarithms of two numbers is equal to the product of the logarithms of the numbers.

\(\log _a(m n)=\log _a m+\log _a n\)

Proof:

Suppose that \(\log _a {~m}={x}\) and \(\log _{{a}} {n}={y}\)
Then \({a}^{{x}}={m}, {a}^{{y}}={n}\)
Hence \({m n}={a}^{{x}} \cdot {a}^{{y}}={a}^{{x}+{y}}\)
It now follows from the definition of logarithms that \(\log _a(m n)=x+y=\log _a m+\log _a n\)

Examples: 

\(
\log _2 5+\log _2 4=\log _2(5 \times 4)=\log _2 20 \\
\log 4 x+\log x=\log (4 x \times x)=\log 4 x^2
\)

 

Division rule

The base remains the same, the logarithm of the quotient of two numbers is equal to the difference of the logarithms of those two numbers.

\(
\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n
\)

Proof:

\(\text { Let } \log _a \mathrm{~m}=\mathrm{x}, \log _{\mathrm{a}} \mathrm{n}=\mathrm{y}\)
Then \(a^x=m, a^y=n\)
Hence \(\frac{m}{n}=\frac{a^x}{a^y}=a^{x-y}\)
Therefore
\(
\log _a\left(\frac{\mathrm{m}}{\mathrm{n}}\right)=\mathrm{x}-\mathrm{y}=\log _{\mathrm{a}} \mathrm{m}-\log _{\mathrm{a}} \mathrm{n}
\)

Example:

\(\log _3\left(\frac{2}{y}\right)=\log _3(2)-\log _3(y)\)

 

Equality Rule of Logarithms

This rule is used while solving the equations involving logarithms. i.e.

\(\log _b a=\log _b c \Rightarrow a=c\)

It is a kind of canceling log from both sides.

Number Raised to Log Property

When a number is raised to log whose base is same as the number, then the result is just the argument of the logarithm. i.e.,

\(a^{\log _a x}=x\)

Here are some examples of this property.

Negative Log Property

The negative logs are of the form – \(\log _b\) a. We can calculate this using the power rule of logarithms.
\(-\log _b a=\log _b a^{-1}=\log _b(1 / a)\)

Thus, \(-\log _b a=\log _b(1 / a)\)

i.e., To convert a negative log into a positive log, we can just take the reciprocal of the argument. Also, to convert a negative log into a positive log, we can take the reciprocal of the base, i.e., \(-\log _b a=\log _{1 / b} a\)

Power Rule

In this rule, the base remains the same, the logarithm of \(m\) to a rational exponent, is equal to the exponent times the logarithm of \(m\).
\(
\log _a\left(m^n\right)=n \log _a m
\)

Proof:

\(
\text { As before, if } \log _a m=x \text {, then } a^x=m
\)
Then \({m}^{{n}}=\left({a}^{{x}}\right)^{{n}}={a}^{{nx}}\)
giving \(\log _a\left({~m}^{{n}}\right)={nx}={n} \log _a {~m}\)

Example:

\(
\log _a\left(2^3\right)=3 \log _a 2
\)

 

Change of base rule

Sometimes, in mathematical calculations involving logarithm, we need to change the base of the logarithm. This rule allows a change of base of the logarithm.

\(
\begin{aligned}
& \log _a b=\frac{\log _c b}{\log _c a} \\
& \log _a b=\frac{1}{\log _b a}
\end{aligned}
\)

Proof:

\(
\text { Let } x=\log _a b
\)
Write in exponent form
\(
a^x=b
\)
Take \(\log _c\) of both sides and evaluate
\(
\begin{aligned}
& \log _c a^x=\log _c b \\
& x \log _c a=\log _c b \\
& x=\frac{\log _c b}{\log _c a} \\
& \log _a b=\frac{\log _c b}{\log _c a}
\end{aligned}
\)
When \(c=b\)
\(
\log _a b=\frac{\log _b b}{\log _b a}=\frac{1}{\log _b a}
\)

Example:

\(\log _b 2=\frac{\log _a 2}{\log _a b}\)

 

Base Switch Rule

\(\log _b(a)=\frac{1}{\log _a(b)}\)

Example:

\(\log _b 8=\frac{1}{\log _8 b}\)

 

Derivative of Log

Derivative of Common Logarithm:
\(
\frac{d}{d x} \log _a(x)=\frac{1}{x \ln (a)}
\)
Derivative of Natural Logarithm:
\(
\frac{d}{d x} \ln (x)=\frac{1}{x}
\)

Proof:

Assume that \(y=\log _a x\). Converting this into the exponential form would give \(a^y=x\). By taking the derivative on both sides with respect to \(x\), we get
\(
d / d x\left(a^y\right)=d / d x \times (x)
\)
By using the chain rule,
\(
\left(a^y \ln a\right) d y / d x=1
\)
\({dy} / {dx}=1 /\left(a^y \ln a\right)\)
But we have \(a^y=x\). Therefore,
\(
d y / d x=1 /(x \ln a)
\)

Similarly we can proof using natural log derivative as well.

It is a logarithm with base ” \(\mathrm{e}\) ” and hence it can be written as \(\ln x=\log _e x\). Now, we have
\(\frac{d}{d x}\left(\log _a x\right)=1 /(x \ln a)\)
Substitute \({a}={e}\) on both sides. Then we get:
\(\frac{d}{d x}\left(\log _e x\right)=1 /(x \ln e)\)
By the properties of natural logarithms,  \(\ln e=1\). So
\(\frac{d}{d x}\left(\log _e x\right)=1 /(x \cdot 1)\)
Thus, \(\frac{d}{d x}\left(\log _e x\right)=1\).
Replacing \(\log _e x\) with \(\ln x\) back, we get \(\frac{d}{d x}(\ln x)=1 / x\).

Example:

\(
\frac{d}{d x}\left(\ln \left(5 x^3-2\right)\right)=\frac{15 x^2}{5 x^3-2}
\)

 

Integral of Log

\(
\int \log x d x=x(\log x-1)+c
\)
or
\(
\int \ln x d x=x(\ln x-1)+c
\)

Proof:

\(
\begin{aligned}
& \int \ln (x) d x \\
& \text { set } \\
& u=\ln (x), \quad d v=d x \\
& \text { then we find } \\
& d u=(1 / x) d x, \quad v=x \\
& \text { substitute } \\
& \int \ln (x) d x=\int u d v \\
& \text { and use integration by parts } \\
& =u v-\int v \text { du } \\
& \quad \operatorname{substitute} u=\ln (x), v=x, \text { and } d u=(1 / x) d x \\
& =\ln (x) x-\int x(1 / x) d x \\
& =\ln (x) x-\int d x \\
& =\ln (x) x-x+c \\
& =x \ln (x)-x+c
\end{aligned}
\)

Example: What is the integration of \(\log \mathrm{x}\)  with base 10?

We can write \(\log x\) with base 10 as \(\log _{10} x=\) \(\left(\log _e x / \log _e 10\right.\) ). Therefore, the integral of \(\log x\) with base 10 is given by,
\(
\begin{aligned}
& \int \log _{10} x d x=\int\left(\log _e x / \log _e 10\right) d x \\
& =\left(1 / \log _e 10\right) \int \log x d x \\
& =\left(1 / \log _e 10\right)[x \log x-x+c]–[\text { Using the formula for the } \\
& \text { integration of } \log x \text { base } e] \\
& =x \log x / \log _e 10-x / \log _e 10+k, \text { where } k=c / \log _e 10 \\
& =x \log _{10} x-x / \log _{10} e+k
\end{aligned}
\)

Expanding Logarithms

Let us expand the logarithm \(\log \left(3 x^2 y^3\right)\).
\(
\begin{aligned}
& \log \left(3 x^2 y^3\right) \\
& =\log (3)+\log \left(x^2\right)+\log \left(y^3\right) \text { (By product rule) } \\
& =\log 3+2 \log x+3 \log y(\text { (By power rule) }
\end{aligned}
\)

Condensing Logarithms

Let us just take the above sum of logarithms and compress it. We should get \(\log \left(3 x^2 y^3\right)\) back.
\(
\begin{aligned}
& \log 3+2 \log x+3 \log y \\
& =\log (3)+\log \left(x^2\right)+\log \left(y^3\right) \text { (By power rule) } \\
& =\log \left(3 x^2 y^3\right) \text { (By product rule) }
\end{aligned}
\)

Important Notes on Logarithms:

• The logarithm of 0 is NOT defined as one number raised to another number never gives 0 as the result.
• An exponential equation is converted into a logarithmic equation and vice versa using \(b^x=a \Leftrightarrow \log_b a=x\).
• A common log is a logarithm with base 10, i.e., \(\log _{10}=\log\).
• A natural log is a logarithm with base e, i.e., \(\log _e=\ln\).
• Logarithms are used to do the most difficult calculations of multiplication and division.