# How do you evaluate log_16(1/4)?

Mar 23, 2016

$x = - \frac{1}{2}$

#### Explanation:

Let ${\log}_{16} \left(\frac{1}{4}\right) = x$, then ${16}^{x} = \frac{1}{4}$ or

${\left({2}^{4}\right)}^{x} = \frac{1}{{2}^{2}}$ or

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

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

Mar 23, 2016

Let's write it in exponential form.

#### Explanation:

${\log}_{a} b = x \to {a}^{x} = b$

${\log}_{16} \left(\frac{1}{4}\right) = x \to {16}^{x} = \frac{1}{4}$

Solve for x by putting everything in the same base.

${\left({2}^{4}\right)}^{x} = \frac{1}{{2}^{2}}$

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

$x = - \frac{1}{2}$

Therefore, ${\log}_{16} \left(\frac{1}{4}\right) = - \frac{1}{2}$

Practice exercises:

1. Evaluate or solve for x.

a). ${\log}_{9} \left(\frac{1}{27}\right)$

b) ${\log}_{x} \left(81\right) = 4$

c) ${\log}_{2 x + 1} 11 x = 2$

Good luck!