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

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Logarythmic
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How do I solve

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

for x?
 
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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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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