Question: > 
> Well the question is to solve for $x$ in $$a^x+b^x=c \tag{a,b,c are constants}$$
> 
> 
> 

Well as of me, I tried to put $\ln{}$ on both sides which does not seem to help. Apart from this I don't seem to have any other way to solve it. 

I turned to WolframAlpha and the solution which was given was :
$$x=\dfrac{-i(2\pi n +\pi)}{\ln{a}-\ln{b}} , {n \in \mathbb{Z}}$$

And if $n=0$ $$x=\dfrac{-i\pi}{\ln{a}-\ln{b}}$$

The only thing I am able to recognize is $e^{i\pi}=1$ ... i guess is used.

Link : <http://goo.gl/da9G01>

So I see that $x$ is *independent* of the constant $c$.

So how to solve it? Please help...Thanks!

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**EDIT** - It seems that WolframAlpha is taking $c=0$ as suggested by Donkey\_2009. And when i write "Solve for x in a^x + b^x = k" it is *not* returning *anything*. So are you guys aware of a method to solve this? Also are you guys aware of any other website/software like WA?

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**P.S.** - Also I see that IF we know the values of $a,b,c$ resulting in something like $2^x+3^x=100$ , then *Newton's method* is the only method I am aware of that can solve it...

**Any other suggestions?**

Answer: 
*See edit below*

I don't think the solution should be independent of $c$. Maybe you typed into Wolfram

$$
a^x+b^x=0
$$

In that case, it's not too hard to see where the solution comes from. Move $b^x$ to one side to get:
\begin{align}
b^x&=-a^x\\
&=(-1)\times a^x\\
&=e^{i\pi}a^x
\end{align}
Then take $\ln$ of both sides:
$$
x\ln(b)=i\pi +x\ln(a)
$$
from which it follows easily that
$$
x=\frac{-i\pi}{\ln(a)-\ln(b)}
$$
The other solutions come from using the fact that we have:
$$
e^{i(2n\pi+\pi)}=(e^{i\pi})^{2n+1}=(-1)^{2n+1}=-1
$$
so we can replace $e^{i\pi}$ with $e^{i(2n\pi+\pi)}$ above and get another solution.

---

If we no longer insist that $c=0$ then we won't get such a nice solution. This is not unusual in mathematics. Even nice polynomial equations like $x^5+x+1=0$ don't have solutions that can be written in terms of nice functions.

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**Edit:** I've just see your link, and it looks as if you've misunderstood Wolfram's answer. You typed in
$$
c=a^x+b^x
$$
and Wolfram Alpha interpreted this as $c$ being a function of $x$, where $a, b$ are constants. The value 
$$
x=\frac{-i\pi}{\ln(a)-\ln(b)}
$$
is listed as a *root* of this equation, i.e., a value of $x$ that makes $c$ equal to $0$. So we are effectively solving
$$
a^x+b^x=0
$$
for $x$. 

As an example, if you typed in
$$
y=x^2-5x+6
$$
then a *root* of this equation would be a value of $x$ such that $x^2-5x+6=0$.
Not a solve, just an idea. This is for $a^x+b^x=0$ by the way.
$$a^x=-b^x$$
$$\log\_ax=-\log\_bx$$
$$\frac{\ln x}{\ln a}=-\frac{\ln x}{\ln b}$$
$$\ln a=- \ln b$$
$$x=\frac{-i \pi}{ln(a)-ln(b)}$$ 
$$x=\frac{-i \pi}{2\ln(a)}$$
$$2x\ln (a)=-i \pi$$
$$e^{2x\ln (a)}=e^{-i \pi}$$
$$e^{2x\ln (a)}=-1$$