How do you solve for log_4 x = -2?

Jul 21, 2018

$x = \frac{1}{16}$

Explanation:

From the definition of logarithm, if ${\log}_{a} b = c$ then $b = {a}^{c}$

Hence as ${\log}_{4} x = - 2$,

we have ${4}^{- 2} = x$

i.e $x = {4}^{- 2} = \frac{1}{4} ^ 2 = \frac{1}{16}$