Answer
Solution set = $\displaystyle \{\frac{e^{3}}{2}\}$.
Work Step by Step
$3+4\ln(2x)=15$
Logarithms are defined only for positive arguments, so any solution must satisfy
$2x \gt 0\Rightarrow\qquad x \gt 0\qquad(*)$
subtract 3 from both sides
$ 4\ln(2x)=12\qquad $... divide with 4
$\ln(2x)=3$
Write the RHS as $\ln($....$)$ using the basic property $\log_{b}b^{x}=x $
$\ln(2x)=\ln e^{3} $
logarithmic functions are one-to-one; if $\log_{b}M=\log_{b}N $, then M=N
$ 2x=e^{3}\qquad $... divide with $2$
$ x=\displaystyle \frac{e^{3}}{2}\approx 10.04$
which satisfies the condition (*), and is a valid solution.
Solution set = $\displaystyle \{\frac{e^{3}}{2}\}$.
