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$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$

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    By inspection, $x = y = -1$ is a solution. Proving that it's the only solution or that other solutions exist is another matter though...2011-11-20
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    If you plot the solution sets you easily see that the above is the only solution to the problem, an analytic way to show that seems hard though.2011-11-20
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    plot($\log_{5}(\frac{7}{10}-2^x)-\log_{2}(\frac{7}{10}-5^x)$),[link](http://i.imgur.com/3rd3c.jpg)2011-11-20
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    It's easy to see that $x=y=-1$ is the only solution where $x=y$. By itself, the equation $5^x+2^y = 2^x+5^y$ is equivalent to $5^x-2^x = 5^y-2^y$. The function $f(x)=5^x-2^x$ is negative for $x<0$ and positive thereafter; also, $f(x)$ is decreasing to the left of $x_0=-\log_{5/2} (\log_2 5)$ and increasing thereafter. So for each $x<0$ ($x\ne x_0$) there is exactly one $y\ne x$ such that $5^x-2^x = 5^y-2^y$. - Having said all that, I don't immediately see how to proceed from here.2011-11-20
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    If you can find $x+y$ then you can find $x$, $y$ by expressing $y$ in terms of $x$ and sum, putting $2^x=a, 5^x=b$ and solving the quadratic equation in $a,b$. Question seems interesting to me so I put a bounty.2011-12-03
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    You can show $f(x)$ (where $f$ is pedja's function, that you get by eliminating $y$) is decreasing in the interval of interest using calculus, and therefore conclude that $x=-1$ is the only solution. I don't see a more elementary (precalculus) argument for why $f$ is decreasing yet, though.2011-12-03
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    In the relation $5^x+2^y=k$ y may be treated as a function of x,considering k a constant.$\frac{dy}{dx}=-\frac{5^x}{2^y}log_{2}5<0$. Similarly in $2^x+5^y=k$, $\frac{dy}{dx}=-\frac{2^x}{5^y}\frac{1}{log_{2}5}<0$. Both are decreasing functions and hence their curves can intersect at only one point.If you get the point x=y=-1 it is the only possible result for x and y.The sum 1/(x+y)=-1/22011-12-04
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    Incidentally $5^x+2^y=7/10=7/(5*2)=(5+2)/5*2=5^{-1}+2^{-1}$. Again,$5^y+2^x=7/10=(5+2)/(5*2)=1/5+1/2=5^{-1}+2^{-1}$. The result x=-1 and y=-1 is an easy prediction.2011-12-04
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    Bounty will be awarded to subjectively cleanest and complete solution, preferably using calculus as less as possible.2011-12-04
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    @MaX: Why did you add the (algebra-precalculus) tag? The OP didn't specify whether calculus should be used.2011-12-05
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    We can see from @Greg's comment that there would be further solutions if we replace $7/10$ by a number between $2^\alpha+5^\alpha\approx0.7565$ and $1$, where $\alpha=-(\log\log5-\log\log2)/(\log5-\log2)\approx-0.9194$. Thus the proof that $x=y=-1$ is the only solution must somehow use the fact that $7/10$ is outside this interval. I don't see how any of the answers given so far do that.2011-12-06
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    @Anamitra I dispute your claim that "both are decreasing functions and hence their curves can intersect at only one point".2011-12-08
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    Given that there is a bifurcation in the number of solutions as the parameter $A$ moves from $0.7$ through $0.9$, and that the exact value of $A$ at the bifurcation is a horribly messy thing that can be computed using my answer (and subsequent comment) below, I suspect that a calculus-free proof of a unique solution for $(x,y)$ is not possible. I'd love to see one though.2011-12-08

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