How do you solve $2^{2 n} \leq \frac{1}{16}$ $2^(2n)<=1/16$ ?

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Answer 1 · 2016-10-13T21:06:59

$n \leq - 2$ $n <=-2$

$log (2^{2 n}) \leq log (\frac{1}{16})$ $log(2^(2n)) <= log(1/16)$

$2 n log 2 \leq log (2^{- 4})$ $2nlog2 <= log(2^(-4))$

$2 n log 2 \leq log (2^{- 4})$ $2nlog2 <=log(2^(-4))$

$2 n log 2 \leq - 4 log 2$ $2nlog2 <=-4log2$

$2 n log 2 + 4 log 2 \leq 0$ $2nlog2 + 4log2 <= 0$

$log 2 (2 n + 4) \leq 0$ $log2(2n + 4) <= 0$

$2 n + 4 \leq 0$ $2n + 4 <= 0$

$n \leq - 2$ $n <= -2$

Hopefully this helps!

Answer 2 · 2016-10-13T21:08:46

$n \leq - 2$ $n<=-2$

$2^{2 n} \leq \frac{1}{16}$ $2^(2n)<=1/16$

$log (2^{2 n}) \leq log (\frac{1}{16})$ $log(2^(2n))<=log(1/16)$

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

$2 n log (2) \leq log (\frac{1}{16})$ $2nlog(2)<=log(1/16)$

$2 n \leq \frac{log (\frac{1}{16})}{log (2)}$ $2n<=log(1/16)/(log(2))$

$2 n \leq - 4$ $2n<=-4$

$n \leq - \frac{4}{2}$ $n<=-4/2$

$n \leq - 2$ $n<=-2$

How do you solve 22n≤1162^(2n)<=1/16?