Solving exponential equation with unknown on both sides

Question

I am having trouble solving the equation

$$3e^{−x+2} = 5e^{x-1}$$

Any help would be appreciated. Thanks.

Answer 1

HINT:

Take log in both sides to find $$\ln3+(-x+2)=\ln5+x-1$$

Answer 2

$$3e^{−x+2} = 5e^{x-1}$$ 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 $\ln\left(\frac{a}{b}\right)=\ln a -\ln b$. $$\therefore x=\frac{1}{2}\Big(3+\ln\left(\frac{3}{5}\right)\Big).$$

Answer 3

we have $\frac{3}{5}=e^{2x-3}$ by the power rule and then $\ln(\frac{3}{5})=2x-3$ can you proceed?

Answer 4

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 $$\frac{3}{5}e^{3} = e^{2x}$$ $$\ln(\frac{3}{5}e^{3}) = \ln(e^{2x})$$ $$\ln(\frac{3}{5}e^{3}) = 2x$$ $$\therefore x = \frac{1}{2}\ln(\frac{3}{5}e^{3})$$

Answer 5

$$3e^{-x+2} = 5e^{x-1}$$ $$\frac{3e^2}{e^x} = \frac{5e^x}{e^1}$$ $$\dots = \dots$$