Answer
$x=\dfrac{33}{2}$
Work Step by Step
Since $\log_b y=x$ implies $y=b^x$, the given equation, $
\log_2(2x-1)=5
$, implies
\begin{align*}\require{cancel}
2x-1&=2^5
\\
2x-1&=32
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
2x&=32+1
\\
2x&=33
\\\\
\dfrac{\cancel{2}x}{\cancel{2}}&=\dfrac{33}{2}
\\\\
x&=\dfrac{33}{2}
.\end{align*}
Hence, the solution to the equation $
\log_2(2x-1)=5
$ is $
x=\dfrac{33}{2}
$.
