Answer
$x=\dfrac{-2\log3-\log2}{3\log2-\log3}\approx-2.946865$
Work Step by Step
$2^{3x+1}=3^{x-2}$
Apply $\log$ to both sides of the equation:
$\log2^{3x+1}=\log3^{x-2}$
The exponents $(3x+1)$ and $(x-2)$ can be taken down to multiply in front of their respective logarithms:
$(3x+1)\log2=(x-2)\log3$
Solve for $x$:
$3x\log2+\log2=x\log3-2\log3$
$3x\log2-x\log3=-2\log3-\log2$
$x(3\log2-\log3)=-2\log3-\log2$
$x=\dfrac{-2\log3-\log2}{3\log2-\log3}\approx-2.946865$
