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How to solve this equation $$\exp\left({\dfrac{b+a}{a}u} \right)+\exp\left({\dfrac{b}{a}u} \right)=w$$ where $u$ is unknown and $w$ is known?

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How would one solve $x^{b/a}(x+1)=w$ in general? I don't think one would; in the very special case $b/a=5$, we need Bring radicals, so the situation looks grim for a general method here. – anon Jul 7 '12 at 14:30
(err, technically that should be $b/a=4$...) – anon Jul 7 '12 at 15:13
If you divide the entire equation by $e^{\frac{b}{a}u}$, you would get $e^u + 1 = \omega\left(e^u\right)^{-\frac{b}{a}}$. Analyzing these two functions would give you an idea of the existence of such solutions, although it is hard to tell which. – Ian Mateus Jul 8 '12 at 2:48
up vote 1 down vote accepted


Letting $x=e^u$ (as anon does in a comment) the equation becomes $x^{b/a}(x+1)=w$ and then $$x^b(x+1)^a=v$$ where $v=w^a$. Even if $a,b$ are positive integers, making this a polynomial equation, there will be no analytic solution if $a,b$ are relatively prime and $a+b\ge5$. So you would have to resort to numerical methods, such as Newton's Method, which generally depend on a knowledge of Calculus.

[Added 9 July] Even in the case $a=2$, $b=1$, the equation $$e^{3u/2}+e^{u/2}=w$$ becomes $$x^3+x=w$$ on setting $x=e^{u/2}$, and while there is a formula for solving that one, an analogue of the quadratic formula but for cubics, that formula is not generally included in the school curriculum nowadays. You won't have any trouble finding it on the web, if you want it - just look for "cubic formula".

I suppose the best way to solve your equation is to find some software designed to do it for you.

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