How to solve for a variable that is only in exponents? Hi there. I've got this equation:
$$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1$$
How can I find the $x$?
Thanks for helping!
 A: Starting with
$$\left(\frac 35\right)^x+\left(\frac 45\right)^x=1$$
we have a non-constant compared with a constant.  Taking the derivative of the LHS, we get
$$y'=\ln\frac 35\left(\frac 35\right)^x+\ln\frac 45\left(\frac 45\right)^x$$
As $\frac 35,\frac 45$ are both less than $1$, so $\ln\frac 35,\ln\frac 45$ are both negative.  But $c^x$ is strictly positive for positive $c$ and real $x$, so $y'\lt 0$ for all $x$.  Therefore the original right-hand side is monotone decreasing for all real $x$, and given the solution $x=2$, it must be the only one.
Note also that while it was relatively straightforward to observe $x=2$ is a solution to this equation, solving this equation for $x$ is non-trivial at best, and at worst requires numerical methods to solve.
In particular, note that
$$\left(\frac 35\right)^x=\left(\frac 45\right)^{x{\ln\frac 35\over\ln\frac45}}$$
which means that the equation to solve can be written as
$$\left(\frac 45\right)^{x{\ln\frac 35\over\ln\frac45}}+\left(\frac 45\right)^x=1$$
or more simply,
$$q^a+q=1$$
where $q=\left(\dfrac 45\right)^x,a={\ln\frac 35\over\ln\frac45}={\ln 3-\ln 5\over\ln 4-\ln 5}\approx 2.2892242269941\dots$  In general, for $u^x+v^x=1$ we would have (w.l.o.g.) $q=u^x,a={\ln v\over\ln u}$, and except for $a\in\Bbb Q$ we would be limited under current knowledge to using only numerical methods.
A: Although $x=2$ is an obvious solution, there is also the problem of showing that there are no other solutions.  To do that, observe that as $x$ gets bigger, $(3/5)^x+(4/5)^x$ gets smaller.
A: Think about the Pythagorean theorem.
