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 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 worksheet related to logarithm and exponential rules and pdf.

Table of Contents

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

Examples based on Standard Logarithm Rules

Download more examples and worksheet on the log

 

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 common logarithm function.

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

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”

Standard Logarithm Rules

Let m and n be arbitrary positive numbers such that $a>0,$$a\ne 1,$ $b>0,$ $b\ne 1$ then 

1. Zero Rule

${{\log }_{a}}1=0$ 

The logarithm of 1 to any positive base is zero. Provided base must not be 1.

Think about  ${{\log }_{1}}1$

This value is not defined because the base becomes zero.     

2. Identity Rule

${{\log }_{a}}b.\,{{\log }_{b}}a=1\,=\,{{\log }_{a}}a={{\log }_{b}}b$ 

The logarithm of any positive number ‘a’ to same base ‘a’ is one. Provided ‘a’ must not be 1.

3. Base Interchange Rule

${{\log }_{a}}b=\frac{1}{{{\log }_{b}}a}$

By taking reciprocal of log we can interchange the base of the log

4. Base Changing Rule

${{\log }_{c}}a={{\log }_{b}}a.\,\,\,{{\log }_{c}}b$ 

or ${{\log }_{c}}a=\frac{{{\log }_{b}}a}{{{\log }_{b}}c}$

we can change the base of log as shown above.

5. Product Rule

${{\log }_{a}}(mn)={{\log }_{a}}m+{{\log }_{a}}n$ 

The logarithm of the product of numbers is equal to the sum of the logarithm of individual numbers

6. Quotient Rule

${{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n$ 


A logarithm of the quotient of two numbers is equal to the difference of logarithm of individual numbers. 

7. Power Rule

${{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m$ 


The logarithm of an exponential number is the exponent times the logarithm of the base.

8. Exponent of Log Rule

${{a}^{{{\log }_{a}}m}}=m$ 


If the base of the exponent is the same as the base of log and are represented in above manner then base and log bot canceled out.

9. Reciprocal Rule

${{\log }_{a}}\left( \frac{1}{n} \right)=-{{\log }_{a}}n$ 


1/n is equal to n raise to power -1, so by using power rule we get the desired result.

10. Base power rule

${{\log }_{{{a}^{\beta }}}}n=\frac{1}{\beta }{{\log }_{a}}n$ 

If the base of logarithm has any exponent then that number divides the whole logarithm as shown above.

11. Mixture of base power and power rule

${{\log }_{{{a}^{\beta }}}}{{n}^{\alpha }}=\frac{\alpha }{\beta }{{\log }_{a}}n$ , $(\beta \ne 0)$ 

It is the mixture of both power rule of log and base power rule of log.

12. Advanced exponent rule

${{a}^{{{\log }_{c}}b}}={{b}^{{{\log }_{c}}a}}\,$, $(a,b,c>0$ and $c\ne 1)$

Just remember advance logarithm rule.

Examples based on Standard Logarithm Rules

Example 5: Simplify ${{\log }_{4}}64$

Solution: Method 1 Using Basic logarithm rule

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

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

or n = 3

Method 2 Using power rule of logarithms ${{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m$ and ${{\log }_{a}}a=1$ by applying above logarithm rules we get,

${{\log }_{4}}64$ = ${{\log }_{4}}{{4}^{3}}$

= $3{{\log }_{4}}4$  = 3

Example 6: Expand ${{\log }_{4}}(64{{x}^{3}}{{y}^{5}})$

Solution: Using product rules of logarithm we get,

${{\log }_{4}}(64{{x}^{3}}{{y}^{5}})={{\log }_{4}}(64)+{{\log }_{4}}{{(x)}^{3}}+{{\log }_{4}}{{(y)}^{5}}$

Now use power rule

$={{\log }_{4}}{{(4)}^{3}}+{{\log }_{4}}{{(x)}^{3}}+{{\log }_{4}}{{(y)}^{5}}$

$=3{{\log }_{4}}(4)+3{{\log }_{4}}(x)+5{{\log }_{4}}(y)$

$=3+3{{\log }_{4}}(x)+5{{\log }_{4}}(y)$ (Using Identity rule of logarithms)

This gives the detailed solution to the given example

Example 7: Expand ${{\log }_{5}}\left( \frac{625{{x}^{4}}}{{{5}^{z}}{{y}^{10}}} \right)$

Solution: Using product rules and quotient rules of logarithm we get,

${{\log }_{5}}\left( \frac{625{{x}^{4}}}{{{5}^{z}}{{y}^{10}}} \right)={{\log }_{5}}(625)+{{\log }_{5}}{{(x)}^{4}}-{{\log }_{5}}{{(5)}^{z}}-{{\log }_{5}}{{(y)}^{10}}$

Now use power rule

$={{\log }_{5}}{{(5)}^{4}}+{{\log }_{5}}{{(x)}^{4}}-{{\log }_{5}}{{(5)}^{z}}-{{\log }_{5}}{{(y)}^{10}}$

$=4{{\log }_{5}}(5)+4{{\log }_{5}}(x)-z{{\log }_{5}}(5)-10{{\log }_{5}}(y)$

$=4+4{{\log }_{5}}(x)-z-10{{\log }_{5}}(y)$ (Using Identity rule of logarithms)

This gives the detailed solution to the given example


Example 8:
 Find Log of $32\sqrt[5]{4}$ to the base $2\sqrt{2}$ is?

Solution: Let x be the required logarithm, then by log rules

${{(2\sqrt{2})}^{x}}=32\sqrt[5]{4}$

${{({{2.2}^{1/2}})}^{x}}={{2}^{5}}{{.2}^{2/5}}$

or ${{2}^{\frac{3x}{2}}}={{2}^{5+\frac{2}{5}}}$

Since the bases are same so by equating the exponents we get

$\frac{3}{2}x=\frac{27}{5}$

Hence x = 18/5

or x = 3.6

Note: This question can also be solved by above mentioned methods 

Brain Teaser: $\log {{x}^{2}}=2\log x$  is it always true? 

Example 9: Find the value of ${{\log }_{100}}(0.00001)$ by using basic logarithm rules.

Solution: Let y = ${{\log }_{100}}(0.00001)$

or ${{100}^{y}}=0.00001$

or ${{10}^{2y}}={{10}^{-5}}$

or 2y = -5

or y = $-\frac{5}{2}$

Example 10: Is the number ${{\log }_{2}}7$ rational?


Solution:
 Suppose, if possible, ${{\log }_{2}}7$is rational, say $p/q$ where p and q are integers, prime to each other.

Then, $\frac{p}{q}={{\log }_{2}}7\,\,\,\,\,$

$\Rightarrow 7={{2}^{p/q}}\,\,\,\,$

Taking exponent ‘q’ on both sides we get

$\Rightarrow {{2}^{p}}={{7}^{q}}$

Which is false since L.H.S is even and R.H.S is odd. Obviously ${{\log }_{2}}7$ is not an integer and hence not a prime number

Example 11: Simplify: $7\log \frac{16}{15}+5\log \frac{25}{24}+3\log \frac{81}{80}$

Solution: Initially simplify each of the logarithmic number using prime factorization and applying Quotient rule and power rule

So we get, 7 (log16 – log (3 ×5)) + 5 (log25 – log(8×3)) + 3 (log81 – log(16×5))

= 7 log16 – 7 log3– 7 log5  + 5 log25 – 5 log8 – 5 log3 + 3 log81 – 3 log16– 3 log5

= 28 log2 – 7 log3– 7 log5  + 10 log5 – 15 log2 – 5 log3 + 12 log3 – 12 log2 – 3log5

= 28 log2  – 15 log2 – 12 log2  + 10 log5 – 3log5 – 7 log5  + 12 log3  – 7log3 – 5 log3

= log 2

Example 12: If ${{\log }_{30}}3\ $ = a and ${{\log }_{30}}5\ $ = b, find ${{\log }_{30}}8\ $in terms of a and b.

Solution: ${{\log }_{30}}8\ \,=\,\ {{\log }_{30}}{{2}^{3}}\ =\ \,3\ {{\log }_{30}}2$

Hence, we must find ${{\log }_{30}}2$.

$\therefore {{\log }_{30}}\ \frac{30}{2}\,\ =\,\ {{\log }_{30}}\ \left( 3\,\ \times \,\ 5 \right)\ \,=\ \,{{\log }_{30}}3\ \,+\,\ {{\log }_{30}}5$

i.e., $\ {{\log }_{30}}30\ -\,\ {{\log }_{30}}2\ \,=\ \,a\ \,+\ \,b$

$1\ \,-{{\log }_{30}}2\ \,=\,\ a\,\ +\,\ b$

or ${{\log }_{30}}2\ \,=\,\ 1\ \,-\left( a\ \,+\,\ b \right)$

Hence, ${{\log }_{30}}8\ $= 3 (1 – a – b)

Note: The number whose logarithm is to be found (in this case, 8) should be factorized into prime factors. 

Example 13: 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 14: 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.

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