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

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Answer 1 · 2016-10-13T21:06:59

$n <=-2$

$log(2^(2n)) <= log(1/16)$

$2nlog2 <= log(2^(-4))$

$2nlog2 <=log(2^(-4))$

$2nlog2 <=-4log2$

$2nlog2 + 4log2 <= 0$

$log2(2n + 4) <= 0$

$2n + 4 <= 0$

$n <= -2$

Hopefully this helps!

Answer 2 · 2016-10-13T21:08:46

$n<=-2$

$2^(2n)<=1/16$

$log(2^(2n))<=log(1/16)$

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

$2nlog(2)<=log(1/16)$

$2n<=log(1/16)/(log(2))$

$2n<=-4$

$n<=-4/2$

$n<=-2$

How do you solve 2^(2n)<=1/162^(2n)<=1/16?