## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=-4$
In exponential form, the given logarithmic equation, $\log_2(8-6x)=5 ,$ is equivalent to \begin{array}{l}\require{cancel} 8-6x=2^5 .\end{array} Hence, the value of the variable that satisfies the given equation is \begin{array}{l}\require{cancel} 8-6x=32 \\\\ -6x=32-8 \\\\ -6x=24 \\\\ x=\dfrac{24}{-6} \\\\ x=-4 .\end{array}