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

## Formula logb c = (loga c)/(loga b)

a is the base.
So 0 < a < 1, a > 1.

Domain and base of a logarithm

## Example 1: log2 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.

log2 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: log2 3 = a, log12 18 = ? The base of log2 3 is 2.

So change the bases to 2.

log12 18 = (log2 18)/(log2 12)

Write the prime factorization of 18 and 12.

18 = 2⋅32

12 = 22⋅3

Prime factorizations

log2 2⋅32 = log2 2 + log2 32

log2 22⋅3 = log2 22 + log2 3

Logarithm of a product

log2 32 = 2 log2 3

log2 22 = 2 log2 2

Logarithm of a power

log2 2 = 1

Logarithm of the base

log2 3 = a

Then (given) = (1 + 2⋅a)/(2⋅1 + a).

Arrange the terms.

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

## Example 3: Simplify (log2 27)(log9 16) Change the bases to 2.

log2 27 = (log2 27)/(log2 2)

log9 16 = (log2 16)/(log2 9)

log2 27 = log2 33

log2 2 = 1

log2 16 = log2 24

log2 9 = log2 32

log2 33 = 3 log2 3

log2 24 = 4 log2 2

log2 32 = 2 log2 3

Cancel log2 3 terms.
(dark gray terms)

log2 2 = 1

4⋅1/2 = 2

3⋅2 = 6

So (given) = 6.

## Example 3: Simplify (log2 27)(log9 16), Another Solution Let's solve this example differently
by changing the bases to 3.

log2 27 = (log3 27)/(log3 2)

log9 16 = (log3 16)/(log3 9)

log3 27 = log3 33

log3 16 = log3 24

log3 9 = log3 32

log3 33 = 3 log3 3

log3 24 = 4 log3 2

log3 32 = 2 log3 3

log3 3 = 1

Cancel log3 2 terms.
(dark gray terms)

3⋅1 = 3

4/(2⋅1) = 2

3⋅2 = 6

So (given) = 6.

As you can see,