Indices and Log

$\begin{aligned} 3^{2x} & = 15^{x+1} & \small \color{#3D99F6} \text{Taking natural logarithm on both sides} \\ 2x \ln 3 & = (x+1)\ln 15 \\ 2x \ln 3 - x\ln 15 & = \ln 15 \\ \implies x & = \frac {\color{#3D99F6} \ln 15}{2\ln 3 - \color{#3D99F6} \ln 15} \\ & = \frac {\color{#3D99F6} \ln 3 + \ln 5}{2\ln 3 - \color{#3D99F6} \ln 3 - \ln 5} \\ & = \boxed{\dfrac {\ln 3 + \ln 5}{\ln 3 - \ln 5}} \end{aligned}$

$3^{2x} = 15^{x+1}$

$9^{x} = 15×15^{x}$

$( \frac {9}{15} )^x = 15$

$( \frac {3}{5} )^x = 15$

$x = \log _{ \frac {3}{5} } {15}$

After applying the change of base theorem:

$x = \frac {\ln 15 } { \ln ( \frac {3}{5} ) } = \boxed { \frac {\ln 3 + \ln 5 } { \ln {3} - \ln5 } }$