# Logarithm Examples and Practice Problems

Take a real number x and bx represents an unique real number. If we write a = bx, then the exponent x is the logarithm of a with log base of b and we can write a = bx as
logba = x
The notation x = logba is called Logarithm Notation.
Before goto the example look at this logarithm rules and logarithm calculator.

Example Logarithm Notations:
(i) 3 = log464 is equivalent to 43 = 64
(ii) 1/2 = log93 is equivalent to √9 = 3

Logarithm Examples
1. Change the below lagarithm log255 = 1/2 to exponential form
log255 = 251/2 = 5

2. Change 1/8 = 8-1 to logarithmic notation.
Solution:
1/8 = 8-1 = log88

3. Evaluate the value for log48
Solution:
x = log48
Then we can say
4x = 8 = 23 ,taking square root both side
2x = 23/2
x = 3/2

4. Find the value of x from logx 100 = 2
solution:
logb 1000 = 3
We can write it as,
b3 = 1000
b3 = 103
So from the above equation
b = 10
5. Find the value for log58 + 5(1/1000)
Solution:
log58 + 5(1/1000) = loglog5(8 x 1/1000)
= log5(1/125)
= log5(1/5)3
= log5(5)-3
= -3log5(5)
= -3 x 1
= -3

6. Solve 2log53 X log9x + 1 = log53
Solution:
We can rewrite the above equation in the below format
=> log532 X log9x = log53 - 1
=> log59 X log9x = log53 - log55
=> log59 X log9x = log5(3/5)
By Using the change of base rule in Left side we get
=> log5x = log5(3/5)
The value of X = 3/5

Logarithm Practice Problems:
1. Find Logarithm for following equations
(i) log5125
(ii) log2(2√2)
(iii) log1/327

2. Find the value of x in following equations
(i) logx 0.001 = -2
(ii) √log2x = 3
(iii) 2 log9 x = 1

3. Simplify the following expression into single term
(i) log102 + log1010
(ii) 3log32 + 4 - 8 log33
(iii) log105 + log10 20 - log1024 + log10 25 - 4

4. Solve the following logarithm equation
(i) 3log52 = 2log5x
(ii) xlog168 = -1
(iii) log5(10 + x) = log5(3 + 4x)