# Chapter 2 - Nonlinear Functions - 2.5 Logarithmic Functions - 2.5 Exercises - Page 98: 17

$-4$

#### Work Step by Step

From the definition of logarithms, for $a>0$, $a\neq 1$, and $x > 0$, $y=\log_{a}x$ means $a^{y}=x$. --------- $y=\displaystyle \log_{2}(\frac{1}{16})\ \$ means $\ \ 2^{y}=\displaystyle \frac{1}{16} .$ Since $\displaystyle \frac{1}{16}=16^{-1}=(2^{4})^{-1}=2^{-4},$ $y=-4$

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