# How do you write log_4 (1/16) = -2 into its exponential form?

Feb 13, 2015

You use the definition of logarithm:
${\log}_{a} x = b \implies {a}^{b} = x$

So you can write your expression. taking the base $a$ of the log (in this case is $4$) to the power of $b = - 2$ which is indeed equal to the argument of the log, i.e.:

${4}^{- 2} = \frac{1}{4} ^ 2 = \frac{1}{16}$

In which you use the fact that: ${x}^{-} a = \frac{1}{x} ^ a$

Alternatively you can take all as exponents of $4$:
${4}^{{\log}_{4} \left(\frac{1}{16}\right)} = {4}^{- 2}$
Which is again:
$\frac{1}{16} = \frac{1}{16}$