Chapter 12 - Exponential Functions and Logarithmic Functions - 12.3 Logarithmic Functions - 12.3 Exercise Set - Page 803: 106

$x=-\displaystyle \frac{7}{16}$

For $x\gt 0$ and $a$ a positive constant other than 1, $\log_{a}x$ is the exponent to which $a$ must be raised in order to get $x$. Thus, $\log_{a}x=m$ means $a^{m}=x$ --- By definition, $\log_{8}(2x+1)=-1$ means $8^{-1}=2x+1$ $\displaystyle \frac{1}{8}=2x+1$ $\displaystyle \frac{1}{8}-1=2x$ $\displaystyle \frac{-7}{8}=2x$ $-\displaystyle \frac{7}{16}=x$