Solving Exponential Problems

Given 4*10^(2-x)=3*e^(2x+3) what would be the value of x?

$\begin{array}{|rcll|} \hline 4*10^{2-x}&=&3*e^{2x+3} \\\\ && \boxed{ 10^{2-x}=e^{(2-x)\ln(10)}} \\\\ 4*e^{(2-x)\ln(10)}&=&3*e^{2x+3} \\\\ \dfrac{4}{3} &=& \dfrac{e^{2x+3}} {e^{(2-x)\ln(10)}} \\\\ \dfrac{4}{3} &=& e^{2x+3-(2-x)\ln(10)} \\\\ \dfrac{4}{3} &=& e^{2x+3-2\ln(10)+x\ln(10)} \\\\ \ln \left(\dfrac{4}{3} \right) &=& \ln\left( e^{2x+3-2\ln(10)+x\ln(10)} \right) \\\\ \ln \left(\dfrac{4}{3} \right) &=& 2x+3-2\ln(10)+x\ln(10) \\\\ 2x+x\ln(10) &=& \ln \left(\dfrac{4}{3} \right)+2\ln(10)-3 \\\\ x\Big(2+\ln(10)\Big) &=& \ln \left(\dfrac{4}{3} \right)+\ln(10^2)-3 \\\\ x\Big(2+\ln(10)\Big) &=& \ln \left(\dfrac{4*10^2}{3} \right)-3 \\\\ x\Big(2+\ln(10)\Big) &=& \ln \left(\dfrac{400}{3} \right)-3 \\\\ \mathbf{x} &=& \mathbf{ \dfrac{\ln \left(\dfrac{400}{3} \right)-3} {2+\ln(10)} } \\ \hline \end{array}$