How do I solve this using logarithms?

$3^{x+1}=2^{x+2}\\ (x+1)\ln 3=(x+2)\ln2\\ x\ln 3 - x\ln 2 = 2\ln 2 - \ln 3\\ x = \dfrac{2\ln 2 - \ln 3}{\ln 3 - \ln 2}$

Use the calculator and you get the answer.

3^(x+1)=2^(x+2)

$\begin{array}{|rcll|} \hline 3^{x+1}& = & 2^{x+2} \\ 3^{x+1}& = & 2^{x+1}\cdot 2^1 \quad & | \quad : 2^{x+1} \\ \frac{ 3^{x+1} } {2^{x+1}} & = & 2 \\ (\frac32)^{x+1} & = & 2 \\ (1.5)^{{x+1}} & = & 2 \quad & | \quad \ln \text{both sides} \\ \ln\Big((1.5)^{x+1} \Big) & = & \ln(2) \quad & | \quad \ln(a^b) = b\cdot \ln(a) \\ (x+1)\cdot \ln(1.5) & = & \ln(2) \quad & | \quad : \ln(1.5) \\ x+1& = & \frac{\ln(2)} { \ln(1.5) } \quad & | \quad -1 \\ x & = & \frac{\ln(2)} { \ln(1.5) } -1 \\ x & = & 0.70951129135 \\ \hline \end{array}$