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I tried taking $\log$ on both side but i ended with $\log(a^{x}-b^{x})$ which is difficult to solve. Does anybody has idea how to solve the above equation for $x$.

Nalin Kumar Sharma
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    i think there is no chance to solve this equation for $x$ (with elementary operations) – Dr. Sonnhard Graubner Dec 27 '15 at 10:30
  • What are $a,b,c$? (rational, real, complex?) – flawr Dec 27 '15 at 10:50
  • Care to elaborate on that, @Dr.SonnhardGraubner? – A.P. Dec 27 '15 at 12:16
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    Except for very few specific cases, there is no analytical solutions and only numerical methods would apply. – Claude Leibovici Dec 27 '15 at 12:36
  • Does the Lambert W function apply here? – Simply Beautiful Art Dec 27 '15 at 16:37
  • Similar: http://math.stackexchange.com/questions/1584078/how-to-simplify-a-sum-of-exponential-equation/1585685#1585685 – Simply Beautiful Art Dec 27 '15 at 16:41
  • @ClaudeLeibovici Could you explain why there are no analytical solutions, or at least provide a reference? – A.P. Dec 29 '15 at 10:29
  • @A.P. Don't you think that Simple Art gave the answer writing $0=x^n+ax+c$ ? – Claude Leibovici Dec 29 '15 at 10:34
  • @ClaudeLeibovici No, I don't. I'm fully aware of the Abel–Ruffini theorem: it only tells us that it is unlikely that we can find an algebraic solution to $x^n + ax + c = 0$, but it doesn't say anything about analytic solutions. For example, it seems that quintic equations can be solved analytically (though I don't know the details). Furthermore, I don't know anything about equations of the form $x^\alpha + ax +c = 0$ with $\alpha$ real non-integer, which may or may not be easier to solve. – A.P. Dec 29 '15 at 11:49
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    @A.P. I am sorry to be unable to argue; I am not a mathematician in the real sense of the word. It seems clear that for a few specific values of $\alpha$, the equation reduces to a trivial polynomial with (eventually) explicit solutions. But, for th general case, I don't see anything else numerical methods. Sorry for that; Cheers :-) – Claude Leibovici Dec 29 '15 at 12:25

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I could try to simplify it a little:$$c=a^x-b^x=a^x-a^{x\log_a(b)}$$$$a^x=u$$$$c=u-u^{\log_a(b)}$$$$0=u^{\log_a(b)}-u+c$$From here, you might notice that we can't solve for $u$ because we can't solve the following general problem:$$0=x^n+ax+c$$If we could solve the above problem for any $n$, then we'd probably have things like algebraic solutions to quintic polynomials and such.

In fact, we pretty much can only solve this if:$$0\le n\le4$$And if $n$ is an integer.

We see that $\log_b(a)$ will most likely not be an integer and it will most likely not be between $4,0$.

Simply Beautiful Art
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