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How can we solve this equation for $x$?
$$4e^x e^y = ( 1 + e^x)^2$$

I got stuck with this problem.
It took a lot of time, however, each time unable to solve.

Attempt:
put e^x = u i.e, x = ln(u) $$4u e^y = ( 1 + u)^2$$ $$4u e^y = 1 + 2u + u^2$$ $$0 = u^2 + (2-4e^y).u + 1$$ $$ 2u = -(2-4e^y) +- \quad \sqrt{(2-4e^y)^2 - 4} $$ $$ 2e^x = -(2-4e^y) +- \quad \sqrt{(2-4e^y)^2 - 4} $$ $$ ln2 + x = ln[-(2-4e^y) +- \quad \sqrt{(2-4e^y)^2 - 4}] $$

Then again it stuck!

Any help will be truly appreciated!!

  • 1
    What have you tried so far? Also, some MathJaX formatting would really help here, so we can read it more easily. – 2017-02-07
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    Your subject and question disagree. Is there a $4$ there or not? – 2017-02-07
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    @ThomasAndrews 4 is a constant, we can just ignore it. – 2017-02-07

1 Answers 1

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Hint: Let $u=e^x$, then we get a quadratic equation $4ue^y=(1+u)^2$ which you can solve with the quadratic formula. Then back substitute for $x$.

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    Many thanks, but, when i tried, the quadratic solution is more complicated than the given problem, – 2017-02-07
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    "you can solve with the quadratic...." What word did you intend to use after that? – 2017-02-07
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    @BhishanPoudel The question asks to solve for $x$. So doesn't that suffices? – 2017-02-07
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    @Bhishan now you can simply back substitute. The answer will be extrememly ugly, but it will work – 2017-02-07
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    @Displayname I got it, thanks. – 2017-02-07