Answer
$\dfrac{21}{8}$
Work Step by Step
The domain of the variable requires that $2x+1>0$ and $x-2>0$
Thus:
$2x>-1 \longrightarrow x>-\dfrac{1}{2}$
$x-2 >0 \longrightarrow x>2$
Therefore, $x>2$
Rearranging the equation gives:
$\log(2x+1)-\log(x-2)=1$
Since $\log_a\left(\dfrac{M}{N}\right) = \log_a M-\log_a N$, then
$\therefore \log(2x+1)-\log(x-2) = \log \left(\dfrac{2x+1}{x-2} \right)$
Hence, the equation above is equivalent to:
$\log \left(\dfrac{2x+1}{x-2} \right) = 1$
$\because \log_a{b} = 1 \longrightarrow a^1=b$, then
\begin{align*}
10^1&=\dfrac{2x+1}{x-2}\\\\
10&=\dfrac{2x+1}{x-2}\\\\
2x+1 &= 10(x-2)\\\\
2x+1&=10x-20\\\\
1+20&=10x-2x\\\\
21&=8x\\\\
\frac{21}{8}&=x
\end{align*}
Therefore, $x= \boxed{\dfrac{21}{8}}$.
