# Logarithm Rules and Examples

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## Logarithm Rules and Examples

 Logarithm Rules and Examples

## Logarithm Rules and Examples an Overview

In this article, you will get complete detail and examples of various Logarithm Rules and Exponent Rules and the relation between log and exponent.

It is essential to grasp the relation between exponent and log to completely understand logarithms and its rules and apply them to various questions and examples.

We will start with very basic logarithm and exponential rules and stretch it to high-level examples. There is also a Relation between natural logarithm and common logarithm. Finally, you can also download logarithm rules pdf, examples, and worksheets related to logarithm and exponential rules and pdf.

Logarithm Rules and Logarithm Definition

Basic Logarithm Rules To Remember

Comparison of Exponential Rules and Logarithm Rules

Examples on Basic Exponential and Logarithmic Rules

Logarithmic Function

Natural and Common Logarithms

Rule to Convert Natural Log to the Log base 10

Standard Logarithmic Rules

More Examples based on Standard Logarithm Rules

## Logarithm Rules and Logarithm Definition

“The Logarithm of a given number to a given base is the exponent of the power to which the base must be raised in order to equal the given number.”

If a > 0 and a ≠ 1 then logarithm of a positive number N is defined as the exponent x of that power of ‘a’ which equals N. These rules are also known as Basic logarithm Rules or Log Rules.

## Basic Logarithm Rules To Remember

Iff     $\mathbf{lo}{{\mathbf{g}}_{a}}N=x\,$

Then  ${{a}^{x}}=N$

Also   ${{a}^{\mathbf{lo}{{\mathbf{g}}_{a}}N}}=N$

## Comparison of Exponential Rules and Logarithm Rules

See the following table for the comparison of exponential rules and logarithm rules and memorize it to be comfortable for further logarithm and exponential rules.

 Exponential Form Logarithmic Form ${{a}^{x}}=N$ $\mathbf{lo}{{\mathbf{g}}_{a}}N=x\,$ ${{2}^{3}}=8$ $\mathbf{lo}{{\mathbf{g}}_{2}}8=3\,$ ${{3}^{4}}=81$ $\mathbf{lo}{{\mathbf{g}}_{3}}81=4\,$ ${{5}^{3}}=125$ $\mathbf{lo}{{\mathbf{g}}_{5}}125=3\,$ $1{{0}^{4}}=10000$ $\mathbf{lo}{{\mathbf{g}}_{10}}10000=4\,$ ${{7}^{1}}=7$ $\mathbf{lo}{{\mathbf{g}}_{7}}7=1\,$ ${{5}^{0}}=1$ $\mathbf{lo}{{\mathbf{g}}_{5}}1=0\,$

## Examples on Basic Exponential and Logarithmic Rules

Example 1: Solve for x in logarithmic equation $x=\mathbf{lo}{{\mathbf{g}}_{4}}4$

Solution: Rewriting logarithmic equation $x=\mathbf{lo}{{\mathbf{g}}_{4}}4$ into exponential form

we get ${{4}^{x}}=4$

Now think which value of x satisfy the above exponential equation?

In other words, we want to find the exponent in which 4 be raised in order to get 4.

Well the answer is quite simple, I think you must have guessed

We can rewrite above equation as ${{4}^{x}}={{4}^{1}}$

Since the bases are the same so by comparing the exponents we get x = 1

Example 2: Find the value of log of 64 base 4 or Evaluate ${{\log }_{4}}64$

Solution: Using Basic logarithm rule as explained above example

Let ${{\log }_{4}}64$= n

Then, ${{4}^{n}}=64$

or ${{4}^{n}}={{4}^{3}}$

or n = 3

Example 3: Simplify ${{\log }_{4}}(-4)$

Solution: Using Basic logarithm rule as explained above example

Let ${{\log }_{4}}(-4)$= n

Then, ${{4}^{n}}=-4$

Now think which exponent of 4 gives -4?

Obviously, there is not an exponent of 4 which gives a negative value.

So, actually, there is no real solution to the given equation.

Example 4: Simplify ${{\log }_{4}}(0)$

Solution: Using Basic logarithm rule as explained above example

Let ${{\log }_{4}}(0)$= n

Then, ${{4}^{n}}=0$

Now think which exponent of 4 gives 0?

Obviously, there is not an exponent of 4 which gives zero.

Hence there is no real solution to the given question.

NOTE: We can’t find the log of 0 or negative numbers.

## Logarithmic Function

The function defined by $f(x)={{\log }_{a}}x,$ where a > 0 and a ≠ 1 is called logarithm function.

Its domain is $(0,\infty )$ and range is all real numbers. a is called the base of the logarithm function.

## Natural and Common Logarithms

When the base is ‘e’ then the logarithmic function is called Napierian logarithm function or Natural logarithm function and when the base is 10, then it is called a common logarithm function.

Note: Natural logarithm = ${{\log }_{e}}a$ = lna

## Rule to Convert Natural Log to the Log base 10

To convert Natural logarithm(log base e)  to common logarithm(log base 10) use the following formula

${{\log }_{e}}a={{\log }_{e}}10.{{\log }_{10}}a$

Note:

1. The logarithm of a number is unique i.e. No number can have two different log to a given base.
1. ${{\log }_{2}}7$ can be read as “log of seven base two”

## More Examples

Example 5: If ${{\log }_{7}}2=m,$then find value of ${{\log }_{49}}28$ in terms of ‘m’.

Solution: ${{\log }_{49}}28=\frac{\log 28}{\log 49}$

$=\frac{\log 7+\log 4}{2\log 7}$

=$\frac{\log 7}{2\log 7}+\frac{\log 4}{2\log 7}$

$=\frac{1}{2}+\frac{1}{2}{{\log }_{7}}4$

= $\frac{1}{2}+\frac{1}{2}.2{{\log }_{7}}2$

=$\frac{1}{2}+{{\log }_{7}}2=\frac{1}{2}+m$$=\frac{1+2m}{2}$

Example 6: If ${{\log }_{e}}\left( \frac{a+b}{2} \right)=\frac{1}{2}({{\log }_{e}}a+{{\log }_{e}}b)$, then find relation between a and b.

Solution: ${{\log }_{e}}\left( \frac{a+b}{2} \right)=\frac{1}{2}({{\log }_{e}}a+{{\log }_{e}}b)$

$=\frac{1}{2}{{\log }_{e}}(ab)={{\log }_{e}}\sqrt{ab}$

$\Rightarrow \frac{a+b}{2}=\sqrt{ab}\,\,$

$\Rightarrow a+b=2\sqrt{ab}$

$\Rightarrow {{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}=0\,\,$

$\Rightarrow \sqrt{a}-\sqrt{b}=0$

$\Rightarrow \,\,\,\,a=b$

### Caution to solve logarithm equations

Consider the equation

log (x – 3) = log (2 – 3x)

taking antilog on both sides we get

x – 3 = 2 – 3x

or 4x = 5

or x = $\frac{5}{4}$

But if we substitute $x=\frac{5}{4}$ in the given equation, both the sides are not real. So, actually there is no solution to the given equation.

Hence, initial conditions on the variable i.e. domain must be written before solving a logarithmic equation or inequation or the solution obtained should be checked back in the given equation.

Example 7:  Simplify the following:

(i)      ${{3}^{-\frac{1}{3}{{\log }_{3}}_{^{27}}}}$

(ii)       ${{2}^{{{\log }_{_{2\sqrt{2}}}}27}}$

(iii)      ${{25}^{{{\log }_{5}}\sqrt{17}}}$

(iv)    ${{\log }_{2}}\ \left( {{\log }_{3}}\,81\, \right)$

(v)      ${{81}^{\frac{1}{{{\log }_{5}}3}}}$

(vi)      ${{\log }_{4}}32\ -\ {{\log }_{32}}4$

(vii)   ${{\log }_{3}}\ {{\log }_{2}}\ {{\log }_{\sqrt{3}}}81$

(viii) $\sqrt{{{\left( \frac{1}{27} \right)}^{1-\frac{lo{{g}_{5}}13}{2\,\ lo{{g}_{5}}\ 9}}}}$

(ix)       ${{\log }_{\pi }}\ \tan \left( 0.25\pi \right)$

(x)     log tan 1° + log tan 2° + log tan 3° + …… + log tan 89°

Solution:

(i)      ${{3}^{-\frac{1}{3}{{\log }_{3}}_{^{27}}}}={{3}^{\log }}{{_{3}}^{{{(27)}^{{-1}/{3}\;}}}}\ =\ \,{{27}^{-\frac{1}{3}\ }}\ ,\ \ \text{using}\ \ \ {{a}^{{{\log }_{a}}N}}\ =\,\ N$

$={{\left( {{3}^{3}} \right)}^{-\frac{1}{3}}}\ =\ \,{{3}^{-1}}\ =\,\ \frac{1}{3}$

(ii)     $\text{Now}\,\,\ {{\log }_{2\sqrt{2}}}27\,\ =\,\ {{\log }_{{{2}^{3/2}}}}27\ \,=\ \,\frac{2}{3}\ \,{{\log }_{^{2}}}\ 27\ \,=\,\ {{\log }_{2}}\ {{27}^{\frac{2}{3}}}$

Hence, $\ {{2}^{{{\log }_{2\sqrt{2}}}27}}\ =\ \,{{2}^{_{{{\log }_{_{2}}}\mathop{\left( 27 \right)}^{^{\frac{2}{3}}}\ \ }}}=\,\,{{\left( {{27}^{\frac{1}{3}}} \right)}^{2}}\ =\ \,{{3}^{2}}\ =\ \,9\$

(iii)    ${{25}^{{{\log }_{5}}\sqrt{17}\ }}=\,\ {{\left( {{5}^{2}} \right)}^{{{\log }_{5}}\sqrt{17}}}\ =\ \,{{5}^{2}}{{\ }^{{{\log }_{5}}\sqrt{17}}}=\ \,\ 5{{\ }^{{{\log }_{5}}({{\sqrt{17)}}^{2}}}}=\,\,\ {{\left( \sqrt{17} \right)}^{2}}\ =\ \,17$

(iv)    ${{\log }_{3}}\ 81\ \,=\,\ {{\log }_{3}}\ {{3}^{4}}\ =\,\ 4$

$\therefore \ {{\log }_{2}}\ \left( {{\log }_{3}}81 \right)\ \,=\,\ {{\log }_{2}}4\ \,=\,\ {{\log }_{2}}\ \left( {{2}^{2}} \right)\ \,=\,\ 2$

(v)     $\frac{1}{{{\log }_{5}}3}\ \,=\,\ {{\log }_{3}}5$

$\therefore \ \,\,\,\,\,\,{{81}^{\frac{1}{{{\log }_{5}}3}}}=\ \,{{81}^{{{\log }_{3}}5}}\ =\,\ {{\left( {{3}^{4}} \right)}^{{{\log }_{3}}5}}\,\ =\ \,{{3}^{{{\log }_{3}}{{5}^{4}}}}\ =\ \,{{5}^{4}}\ =\ \,625$

(vi)    Now    ${{\log }_{4}}\ 32\ \,=\,\ \frac{{{\log }_{2}}\ 32}{{{\log }_{2}}\ 4}\,\ =\,\ \frac{{{\log }_{2}}\ {{2}^{5}}}{{{\log }_{2}}\,{{2}^{_{2}}}}\,\ =\,\,\frac{5}{2}\ \ \text{and}\,\ \ \,{{\log }_{32}}\,4\,\,=\,\,\frac{1}{{{\log }_{4}}\,32}\,\,=\,\,\frac{2}{5}$

Hence the given expression $=\ \,\frac{5}{2}\,\ -\,\ \frac{2}{5}\ \,=\ \,\frac{21}{10}$

(vii)   ${{\log }_{\sqrt{3}}}81\ \,=\,\,{{\log }_{{{3}^{1/2}}}}81\ \,=\,\ 2\ {{\log }_{3}}\ 81\,\,=\ \,2\ {{\log }_{3}}\ {{3}^{4}}\ =\ \,2\ \,\times \ \,4\ \,=\ \,8$

$\therefore \ {{\log }_{2}}\ {{\log }_{\sqrt{3}}}81\ \,=\,\ {{\log }_{2}}8\ \,=\,\ {{\log }_{2}}\ {{2}^{3}}\ =\ 3$

$\therefore \ {{\log }_{3}}\,{{\log }_{2}}\ {{\log }_{\sqrt{3}}}81\,\,=\,\,{{\log }_{3}}\,3\,\,=\,\,1$

(viii)  The expression $=\ \left( \frac{1}{\sqrt{27}} \right)\ {{(27)}^{\frac{1}{8}\,{{\log }_{9}}\ 13}}$

$=\frac{1}{3\sqrt{3}}{{3}^{\frac{3}{8}{{\log }_{9}}\ 13}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{since}\,\,\ {{\log }_{9}}\ 13\ \,=\,\ {{\log }_{3}}^{2}13\ \,=\,\ \frac{1}{2}\ {{\log }_{3}}\ 13)$

$=\ \frac{1}{3\sqrt{3}}{{3}^{\frac{3}{16}{{\log }_{3}}\ 13}}\ =\ \,\frac{1}{3\ \sqrt{3}}\ {{3}^{{{\log }_{3}}}}\ \left( {{13}^{\frac{3}{16}}} \right)\ \,\frac{{{13}^{\frac{3}{16}}}}{3\sqrt{3}}\ \,=\ \,\frac{1}{3\sqrt{3}}\ \,\sqrt[16]{2197}\,$

(ix)    $\tan \ \,\left( 0.25\pi \right)\ \,=\,\ \tan \,\ \frac{\pi }{4}\,=\,\ 1$

$\therefore \ {{\log }_{\pi }}\ \tan \,\ \left( 0.25\pi \right)\ \,=\,\ {{\log }_{\pi }}\ \,1\,\ =\,\ 0$

(x)     = log { tan 1° × tan 2° ….. tan 44° × tan 45° × tan 46° ….. tan 89°}

using log a + log b = log ab

= log { tan 1° × tan 2° ….. tan 44° × tan 45° × cot 44° ….. cot 1°}

since tan (90 – q) = cot q

= log 1  (since the factors like tan 1° and cot 1° are reciprocals).

= 0

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