+118608 When you are looking for a power you have to use logs (Unless you can get the bases the same)
$$\begin{array}{rlll}
3^{2x-1}&=&4^{x+2}\qquad&\\\\
log(3^{2x-1})&=&log(4^{x+2})&\\\\
(2x-1)log(3)&=&(x+2)log(4)&\\\\
2x-1&=&\left(\frac{log4}{log3}\right)(x+2)&\\\\
2x-1&=&\left(\frac{log4}{log3}\right)x+\left(\frac{log4}{log3}\right) 2&\\\\
2x-\left(\frac{log4}{log3}\right)x&=&\left(\frac{log4}{log3}\right) 2+1&\\\\
x\left(2-\left(\frac{log4}{log3}\right)\right)&=&\left(\frac{log4}{log3}\right) 2+1&\\\\
x\left(2-\frac{log4}{log3}\right)&=&\left(\frac{2log4}{log3}\right) +1&\\\\
x&=&\frac{\left(\frac{2log4}{log3}\right) +1}{\left(2-\frac{log4}{log3}\right)}&\\\\
\end{array}$$
$${\frac{\left(\left({\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{log4}}}{{\mathtt{log3}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{log4}}}{{\mathtt{log3}}}}\right)\right)}} = {\mathtt{4.773\: \!778\: \!228\: \!378\: \!637\: \!1}}$$
+118608 When you are looking for a power you have to use logs (Unless you can get the bases the same)
$$\begin{array}{rlll}
3^{2x-1}&=&4^{x+2}\qquad&\\\\
log(3^{2x-1})&=&log(4^{x+2})&\\\\
(2x-1)log(3)&=&(x+2)log(4)&\\\\
2x-1&=&\left(\frac{log4}{log3}\right)(x+2)&\\\\
2x-1&=&\left(\frac{log4}{log3}\right)x+\left(\frac{log4}{log3}\right) 2&\\\\
2x-\left(\frac{log4}{log3}\right)x&=&\left(\frac{log4}{log3}\right) 2+1&\\\\
x\left(2-\left(\frac{log4}{log3}\right)\right)&=&\left(\frac{log4}{log3}\right) 2+1&\\\\
x\left(2-\frac{log4}{log3}\right)&=&\left(\frac{2log4}{log3}\right) +1&\\\\
x&=&\frac{\left(\frac{2log4}{log3}\right) +1}{\left(2-\frac{log4}{log3}\right)}&\\\\
\end{array}$$
$${\frac{\left(\left({\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{log4}}}{{\mathtt{log3}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{log4}}}{{\mathtt{log3}}}}\right)\right)}} = {\mathtt{4.773\: \!778\: \!228\: \!378\: \!637\: \!1}}$$
