# How do you solve 2^(2n)<=1/16?

Oct 13, 2016

$n \le - 2$

#### Explanation:

$\log \left({2}^{2 n}\right) \le \log \left(\frac{1}{16}\right)$

$2 n \log 2 \le \log \left({2}^{- 4}\right)$

$2 n \log 2 \le \log \left({2}^{- 4}\right)$

$2 n \log 2 \le - 4 \log 2$

$2 n \log 2 + 4 \log 2 \le 0$

$\log 2 \left(2 n + 4\right) \le 0$

$2 n + 4 \le 0$

$n \le - 2$

Hopefully this helps!

Oct 13, 2016

$n \le - 2$

#### Explanation:

${2}^{2 n} \le \frac{1}{16}$

In order to solve this inequality, we have to use [logarithms].(https://www.khanacademy.org/math/algebra-home/alg-exp-and-log/alg-introduction-to-logarithms/v/logarithms)

$\log \left({2}^{2 n}\right) \le \log \left(\frac{1}{16}\right)$

Using the power rule $-$ (${\log}_{b} \left({x}^{y}\right) = y {\log}_{b} \left(x\right)$) $-$ we can rewrite this inequality as:

$2 n \log \left(2\right) \le \log \left(\frac{1}{16}\right)$

$2 n \le \log \frac{\frac{1}{16}}{\log \left(2\right)}$

$2 n \le - 4$

$n \le - \frac{4}{2}$

$n \le - 2$