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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!

Patrick M
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techrod
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    http://math.stackexchange.com/questions/565163/displaystyle-3x4x-5x – lab bhattacharjee Apr 24 '14 at 15:58
  • 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

3 Answers3

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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.

abiessu
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  • +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
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    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
  • @abiessu Thank you very much!+1 That does the trick! – techrod Apr 24 '14 at 19:30
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    @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
  • I don't think you need derivatives to know that this decreases monotonically. – Michael Hardy Apr 25 '14 at 02:24
  • @MichaelHardy: I believe you are correct, but having applied the derivative mechanism, I think I'll stick with it unless I find a more elegant way together with a way to find that $2$ is a solution apart from observation – abiessu Apr 25 '14 at 04:01
  • @MichaelHardy: Cool. It might be worth noting that in your question. For two reasons. Firstly I was initially suprised and confused by where the x=2 came from. Secondly my reading of the question is that the title is about solving for a variable in exponents and feels like a question about the general case. All the work you have done is obviously relevant for proving there is only one solution but it might be worth noting that there you'd have to use numerical techniques to get the actual answer. Just my thoughts. :) – Chris Apr 25 '14 at 09:05
  • @Chris : You addressed a comment to me, but the comment's content makes it look as if it was intended for someone else --- maybe for the person who posted the question? – Michael Hardy Apr 25 '14 at 14:55
  • @MichaelHardy: sorry, I meant abiessu whose answer it was. I'm not sure how I got confused about it. :) – Chris Apr 25 '14 at 15:41
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    @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
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    @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
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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.

Michael Hardy
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  • yeah, you can verify by plotting! – MattAllegro Apr 24 '14 at 16:30
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    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
  • Damn I made a bad example. – user137794 Apr 24 '14 at 16:54
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    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
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    @MichaelHardy: isn't that $\to\infty$ as $x\to-\infty$? – abiessu Apr 24 '14 at 17:28
  • @abiessu : Yes. A pretty bad typo. – Michael Hardy Apr 25 '14 at 02:23
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Think about the Pythagorean theorem.

Umberto P.
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