# Change of Base Formula

How to use the change of base formula to solve logarithmic problems: formula, examples, and their solutions.

## Formula

log_{b} *c* = (log_{a} *c*)/(log_{a} *b*)*a* is the base.

So 0 < *a* < 1, *a* > 1.

Domain and base of a logarithm

## Example 1: log_{2} 70 = ? (log 2 = 0.301, log 3 = 0.477)

The given conditions (log 2, log 3) are common logs.

So change the bases to 10.

log_{2} 70 = (log 70)/(log 2)

Common logarithm

70 = 7⋅10

log 7⋅10 = log 7 + log 10

Logarithm of a product

log 7 = 0.845

log 10 = 1

log 2 = 0.301

So (given) = (0.845 + 1)/0.301.

0.845 + 1 = 1.845

Move the decimal points of both numbers

3 digits to the right.

(= Multiply 1000

to both of the numerator and the denominator.)

1845/301 = 6.129...

So (given) = 6.129....

But the given conditions (log 2 = 0.301, log 3 = 0.477)

have 3 significant digits.

So write the value with 3 significant digits

by rounding 6.129... to the nearest hundredth.

9 is between 5 and 9.

Then increase the hundredth digit 2 to 3

6.129... → 6.13.

So (given) = 6.13.

## Example 2: log_{2} 3 = *a*, log_{12} 18 = ?

The base of log_{2} 3 is 2.

So change the bases to 2.

log_{12} 18 = (log_{2} 18)/(log_{2} 12)

Write the prime factorization of 18 and 12.

18 = 2⋅3^{2}

12 = 2^{2}⋅3

Prime factorizations

log_{2} 2⋅3^{2} = log_{2} 2 + log_{2} 3^{2}

log_{2} 2^{2}⋅3 = log_{2} 2^{2} + log_{2} 3

Logarithm of a product

log_{2} 3^{2} = 2 log_{2} 3

log_{2} 2^{2} = 2 log_{2} 2

Logarithm of a power

log_{2} 2 = 1

Logarithm of the base

log_{2} 3 = *a*

Then (given) = (1 + 2⋅*a*)/(2⋅1 + *a*).

Arrange the terms.

Then (given) = (2*a* + 1)/(*a* + 2).

## Example 3: Simplify (log_{2} 27)(log_{9} 16)

Change the bases to 2.

log_{2} 27 = (log_{2} 27)/(log_{2} 2)

log_{9} 16 = (log_{2} 16)/(log_{2} 9)

log_{2} 27 = log_{2} 3^{3}

log_{2} 2 = 1

log_{2} 16 = log_{2} 2^{4}

log_{2} 9 = log_{2} 3^{2}

log_{2} 3^{3} = 3 log_{2} 3

log_{2} 2^{4} = 4 log_{2} 2

log_{2} 3^{2} = 2 log_{2} 3

Cancel log_{2} 3 terms.

(dark gray terms)

log_{2} 2 = 1

4⋅1/2 = 2

3⋅2 = 6

So (given) = 6.

## Example 3: Simplify (log_{2} 27)(log_{9} 16), Another Solution

Let's solve this example differently

by changing the bases to 3.

log_{2} 27 = (log_{3} 27)/(log_{3} 2)

log_{9} 16 = (log_{3} 16)/(log_{3} 9)

log_{3} 27 = log_{3} 3^{3}

log_{3} 16 = log_{3} 2^{4}

log_{3} 9 = log_{3} 3^{2}

log_{3} 3^{3} = 3 log_{3} 3

log_{3} 2^{4} = 4 log_{3} 2

log_{3} 3^{2} = 2 log_{3} 3

log_{3} 3 = 1

Cancel log_{3} 2 terms.

(dark gray terms)

3⋅1 = 3

4/(2⋅1) = 2

3⋅2 = 6

So (given) = 6.

As you can see,

you got the same answer.