How do you solve ${log}_{8} z \leq - 2$ $log_8 z<=-2$ ?

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How do you solve ${log}_{8} z \leq - 2$ $log_8 z<=-2$ ?

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Answer 1 · 2016-12-20T23:56:42

I got $z \leq \frac{1}{64}$ $z<=1/64$

Explanation:

We can use the definition of logarithm and write:
$z \leq 8^{- 2}$ $z<=8^-2$
$z \leq \frac{1}{8^{2}}$ $z<=1/8^2$
$z \leq \frac{1}{64}$ $z<=1/64$
$z \leq 0.015625$ $z<=0.015625$

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How do you solve ${log}_{8} z \leq - 2$ $log_8 z<=-2$ ?

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How do you solve ${log}_{8} z \leq - 2$ $log_8 z<=-2$ ?