Answer
$\left\{\dfrac{3}{8}\right\}$
Work Step by Step
Using the properties of equality, the given equation, $
\log(2x+3)=1+\log x
$, is equivalent to
\begin{align*}\require{cancel}
\log(2x+3)-\log x&=1
.\end{align*}
Using the properties of logarithms, the equation above is equivalent to
\begin{align*}\require{cancel}
\log\dfrac{2x+3}{x}&=1
&(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y)
\\\\
\log_{10}\dfrac{2x+3}{x}&=1
&(\text{use }\log x=\log_{10} x)
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, the equation above implies
\begin{align*}\require{cancel}
\dfrac{2x+3}{x}&=10^1
\\\\
\dfrac{2x+3}{x}&=10
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
\cancel x\cdot\dfrac{2x+3}{\cancel x}&=10\cdot x
\\\\
2x+3&=10x
\\
3&=10x-2x
\\
3&=8x
\\\\
\dfrac{3}{8}&=\dfrac{\cancel8x}{\cancel8}
\\\\
\dfrac{3}{8}&=x
.\end{align*}
Hence, the solution set of the equation $
\log(2x+3)=1+\log x
$ is $
\left\{\dfrac{3}{8}\right\}
$.
