# Solving for $x$ : $a^x+b^x=c$

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.

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

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?

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?

• Doesn't seem like the solution is, in general, constructible. – mniip Jul 12 '15 at 16:24
• Has Wolfram ever been wrong before? – user3728501 Jul 12 '15 at 16:36
• @user3728501 I guess once/twice. But ... its rare. – NeilRoy Jul 12 '15 at 16:52
• I have tested the page with much more simpler equations like $m=a\cdot x$. It seems that the web page is not supposed to solve equations. – ajotatxe Jul 12 '15 at 17:08
• @ajotatxe WolframAlpha is VERY much supposed to be able to solve equations....Try this : Solve for x in m=ax [Link: goo.gl/j5HAEm] – NeilRoy Jul 12 '15 at 17:11

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.

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 $x^2-5x+6=0$.

• Check the link I gave :) – NeilRoy Jul 12 '15 at 17:07
• Downvoters - any improvements I can make? :) – John Gowers Jul 12 '15 at 17:11
• Do you know of any other websites/softwares which functions similarly to WA? – NeilRoy Jul 12 '15 at 17:15
• Why do you ask? Are you unhappy with the answer you got from them? As I have tried to explain, it is completely correct. – John Gowers Jul 12 '15 at 17:28

Not a solve, just an idea. This is for $a^x+b^x=0$ by the way. $$a^x=-b^x$$ $$\log_ax=-\log_bx$$ $$\frac{\ln x}{\ln a}=-\frac{\ln x}{\ln b}$$ $$\ln a=- \ln b$$ $$x=\frac{-i \pi}{ln(a)-ln(b)}$$ $$x=\frac{-i \pi}{2\ln(a)}$$ $$2x\ln (a)=-i \pi$$ $$e^{2x\ln (a)}=e^{-i \pi}$$ $$e^{2x\ln (a)}=-1$$