Solving of the equation $e^{-x\pi}-2e^{-2x\pi}=0$

Question

If got problems with the answer for $x$ for the equation:

$$e^{-x\pi}-2e^{-2x\pi}=0$$

I don't really know how to start. I tried to substitute $e^x$ but that didn't really work...

I would be happy if someone could provide a little hint for me! Thanks!

Answer 1

Hint: introduce a new variable $y:= e^{-\pi x}$. Also keep in mind that the exponential function is positive everywhere.

Answer 2

Let $y=e^{-x\pi}$ so $y=2y^2$. Therefore...

Answer 3

$$e^{ -xπ }-2e^{ -2xπ }=0\\ e^{ -xπ }=2e^{ -2xπ }\\ e^{ xπ }=2\\ x=\frac { \ln { 2 } }{ \pi } $$

Answer 4

Come on,

$$e^{-x\pi}=2e^{-2x\pi}\iff -x\pi=\ln2-2x\pi.$$