# 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!

-
This problem has an application. Suppose it were observed that 10% of the books in a library get 70% of the circulation. Now find the value of $x$ for which $0.1^x+0.7^x = 1$, and let $p=0.1^x$. If the proportion of books whose circulation frequency exceeds $f$ is $(f_0/f)^\alpha$ for some $f_0>0$ and some $\alpha>1$ (i.e. a Pareto distribution, which has been used to model this situation) then $p\cdot100\%$ of the books get $(1-p)\cdot100\%$ of the circulation, and $\alpha=(\log p)/(\log(p/(1-p))$. ${}\qquad{}$ – Michael Hardy Apr 25 '14 at 15:50

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.

-
+1 for giving some solid math behind the idea of "plot it and observe that it only goes in one direction" – Tim S. Apr 24 '14 at 17:27
How do you get the solution x=2? You might be able to guess it in this case but you might not have so much luck for different solutions so how would you then solve it? – Chris Apr 24 '14 at 18:34
@Chris: from everything I can see, the only way to get this solution (apart from seeing the Pythagorean Triple $(3,4,5)$) is to use a numerical method. – abiessu Apr 24 '14 at 21:55
@Chris: I haven't given up looking for an "elegant" way to show that $x=2$ is a solution, but when I do (or when I find one) I'll update this answer to include my findings. – abiessu Apr 25 '14 at 15:43
@Chris: with the current update to my answer, I am officially giving up on any elegant ways to show that $x=2$ is a solution. – abiessu Apr 25 '14 at 22:03

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.

-
yeah, you can verify by plotting! – MattAllegro Apr 24 '14 at 16:30
Is there a more algebraic approach to solving this problem? Say if the numbers were slightly changed $$\left(\frac37\right)^x+\left(\frac47\right)^x=1$$ – user137794 Apr 24 '14 at 16:35
@user137794: taking the same approach, it is obvious that $x=1$ is a solution to $$\left(\frac37\right)^x+\left(\frac47\right)^x=1$$and the same logic applies, i.e., that the LHS is monotone decreasing for all real $x$, therefore if a solution exists, it is the only one. – abiessu Apr 24 '14 at 16:42
So the question is how to solve $a^x+b^x=1$ for $x$, when $0<a,b<1$. If there's a closed form it will involve logarithms. However, since $a^x+b^x\to0$ as $x\to\infty$ and $\to\infty$ as $x\to0$, and moreover is a continuous function of $x$, there must exist exactly one solution, and it can be found numerically if not in closed form. – Michael Hardy Apr 24 '14 at 17:05
@MichaelHardy: isn't that $\to\infty$ as $x\to-\infty$? – abiessu Apr 24 '14 at 17:28