Identity

$lo g a^{n} = n lo g a$

Algebra • Logarithm • Laws of Logarithms

Formula

lo g (a^{n}) = n lo g a

For

a > 0

and any real

n

, in any valid base,

lo g (a^{n}) = n lo g a

How to use it

1Take logarithms of both sides of an exponential equation to bring the unknown down out of the exponent.
2Evaluate a large power from a tabulated logarithm: $lo g_{10} 2^{50} = 50 lo g_{10} 2$ .
3It holds for negative and fractional $n$ too, so $lo g a = \frac{1}{2} lo g a$ .

Given

lo g_{10} 2 = 0.3010

, find

lo g_{10} 2^{20}

Step-by-step

1
$lo g_{10} 2^{20} = 20 lo g_{10} 2$
Why: Bring the exponent down in front of the logarithm.
2
$= 20 \times 0.3010 = 6.020$

Answer

6.020

Watch out

lo g (a^{n})

n lo g a

, but

(lo g a)^{n}

is not. There is no law at all for

lo g (a + b)

: only products, quotients and powers can be broken up.

Builds on

Also known as

log power rule, bring the exponent down, logarithm of a power, take logs of both sides