Solving exponential equation with unknown on both sides I am having trouble solving the equation
$$3e^{−x+2} = 5e^{x-1}$$
Any help would be appreciated. Thanks.
 A: 
$$3e^{−x+2} = 5e^{x-1}$$
  Taking the natural $\log$ of both sides gives:
  $$\ln (3e^{−x+2}) = \ln(5e^{x-1})$$
  $$\ln3+(2-x) = \ln5+(x-1) ~~~(\text{ since } \ln(ab)=\ln a + \ln b).$$
  So $$\ln 3 - \ln 5+3=2x.$$
  But $\ln\left(\frac{a}{b}\right)=\ln a -\ln b$. 
  $$\therefore x=\frac{1}{2}\Big(3+\ln\left(\frac{3}{5}\right)\Big).$$

A: we have $\frac{3}{5}=e^{2x-3}$ by the power rule and then $\ln(\frac{3}{5})=2x-3$ can you proceed?
A: HINT:
Take log in both sides to find $$\ln3+(-x+2)=\ln5+x-1$$
A: You can also make the equation much simpler just by remembering your properties of exponents:
$$3e^{−x+2} = 5e^{x-1}$$
$$3e^{−x}e^{2} = 5e^{x}e^{-1}$$
multiply across by e
$$3e^{-x}e^{3}=5e^{x}$$
multiply across by $e^{x}$ and divide by $5$, take the natural log of both sides...
$$3e^{3} = 5e^{2x}$$
Now you've got the $x$ on one side, :) rest ain't so bad
$$\frac{3}{5}e^{3} = e^{2x}$$
$$\ln(\frac{3}{5}e^{3}) = \ln(e^{2x})$$
$$\ln(\frac{3}{5}e^{3}) = 2x$$
$$\therefore x = \frac{1}{2}\ln(\frac{3}{5}e^{3})$$
A: $$3e^{-x+2} = 5e^{x-1}$$
$$\frac{3e^2}{e^x} = \frac{5e^x}{e^1}$$
$$\dots = \dots$$
