Chapter 5 - Exponential and Logarithmic Functions - Chapter Review - Review Exercises - Page 346: 34

$x=\frac{-16}{9}$.

Note that: $8^{6+3x}=(2^3)^{6+3x}=2^{18+9x}$ $4=2^2$ Thus, the given equation is equivalent to: $2^{18+9x}=2^2$. The base is the same (and not 1), hence they are equal if the exponents are equal. Therefore \begin{align*} 18+9x&=2\\ 9x&=2-18\\ 9x&=-16\\ \frac{9x}{9}&=\frac{-16}{9}\\ x&=-\frac{16}{9} \end{align*}