Take a real number x and b^{x} represents an unique real number. If we write a = b^{x}, then the exponent x is the logarithm of a with log base of b and we can write a = b^{x} as

log_{b}a = x

The notation x = log_{b}a is called **Logarithm Notation**.

Before goto the example look at this *logarithm rules* and logarithm calculator.

**Example Logarithm Notations:**

(i) 3 = log_{4}64 is equivalent to 4^{3} = 64

(ii) 1/2 = log_{9}3 is equivalent to √9 = 3

**Logarithm Examples**

1. Change the below lagarithm log_{25}5 = 1/2 to exponential form

log_{25}5 = 25^{1/2} = 5

2. Change 1/8 = 8^{-1} to logarithmic notation.

Solution:

1/8 = 8^{-1} = log_{8}8

3. Evaluate the value for log_{4}8

Solution:

x = log_{4}8

Then we can say

4^{x} = 8 = 2^{3} ,taking square root both side

2^{x} = 2^{3/2}

x = 3/2

4. Find the value of x from log_{x} 100 = 2

solution:

log_{b} 1000 = 3

We can write it as,

b^{3} = 1000

b^{3} = 10^{3}

So from the above equation

b = 10

5. Find the value for log_{5}8 + _{5}(1/1000)

Solution:

log_{5}8 + _{5}(1/1000) = loglog_{5}(8 x 1/1000)

= log_{5}(1/125)

= log_{5}(1/5)^{3}

= log_{5}(5)^{-3}

= -3log_{5}(5)

= -3 x 1

= -3

6. Solve 2log_{5}3 X log_{9}x + 1 = log_{5}3

Solution:

We can rewrite the above equation in the below format

=> log_{5}3^{2} X log_{9}x = log_{5}3 - 1

=> log_{5}9 X log_{9}x = log_{5}3 - log_{5}5

=> log_{5}9 X log_{9}x = log_{5}(3/5)

By Using the change of base rule in Left side we get

=> log_{5}x = log_{5}(3/5)

The value of X = 3/5

**Logarithm Practice Problems:**

1. Find Logarithm for following equations

(i) log_{5}125

(ii) log_{2}(2√2)

(iii) log_{1/3}27

2. Find the value of x in following equations

(i) log_{x} 0.001 = -2

(ii) √log_{2}x = 3

(iii) 2 log_{9} x = 1

3. Simplify the following expression into single term

(i) log_{10}2 + log_{10}10

(ii) 3log_{3}2 + 4 - 8 log_{3}3

(iii) log_{10}5 + log_{10} 20 - log_{10}24 + log_{10} 25 - 4

4. Solve the following logarithm equation

(i) 3log_{5}2 = 2log_{5}x

(ii) xlog_{16}8 = -1

(iii) log_{5}(10 + x) = log_{5}(3 + 4x)

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