Solving for x in D(e^{-2ax}-2e^{-ax})-E=0

[Logarythmic]

How do I solve

$$D(e^{-2ax}-2e^{-ax})-E=0$$

for x?

 

[HallsofIvy]

D and E are constants?

$e^{-2ax}= (e^{ax})^2$. Let $y= e^{-ax}$ and your equation becomes $D(y^2- 2y)- E= 0$. Solve that equation, using the quadratic formula perhaps, and then $x= -ln(y)/a$.