TITLE:  $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$
QUESTION [17 upvotes]: $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$
Work out the values of $\frac{1}{x+y}$

REPLY [9 votes]: As has been noted in other responses, we can eliminate $y$ from the system
$$\begin{align*}
5^x+2^y&=0.7\\
2^x+5^y&=0.7
\end{align*}$$
to obtain the equation in $x$ $$\frac{\ln(0.7-2^x)}{\ln5}=\frac{\ln(0.7-5^x)}{\ln2}$$  This equation has the already observed (in comments) solution $x=-1$.  Can there be other solutions?  We will show that there cannot be any.  The proof below is elementary but does use standard calculus methods twice.  It also would fail if $0.7$ were replaced by something like $0.9$, addressing joriki's concern in comments.
Consider the function $f$ with $$f(x)=\frac{\ln(0.7-2^x)}{\ln5}-\frac{\ln(0.7-5^x)}{\ln2}$$ We find that $$f'(x)=\frac{-2^x\ln2}{\ln5(0.7-2^x)}+\frac{5^x\ln5}{\ln2(0.7-5^x)}$$ Note that $f'$ exists wherever $f$ was defined. We will show that $f'(x)\neq0$ for any $x$, implying that there can be no more than one solution to $f(x)=0$. 
If $f'(x)$ were to equal $0$, then that equation is easily rearranged to $$\frac{\ln5}{\ln2}\left(\frac{0.7}{2^x}-1\right)-\frac{\ln2}{\ln5}\left(\frac{0.7}{5^x}-1\right)=0$$ We are attempting to show that this equation has no solutions.  We have put the equation in this form because it leaves the left side very easily differentiable. If we let $$g(x)=\frac{\ln5}{\ln2}\left(\frac{0.7}{2^x}-1\right)-\frac{\ln2}{\ln5}\left(\frac{0.7}{5^x}-1\right)$$ then we can optimize $g$.  We have that $$g'(x)=-\ln5\frac{0.7}{2^x}+\ln2\frac{0.7}{5^x}$$ It is an easy matter to solve for $g'(x_c)=0$: $$x_c=\frac{\ln\left(\frac{\ln5}{\ln2}\right)}{\ln\left(\frac25\right)}<0$$ So $g$ has precisely one critical point.  $g''(x)$ is also easy to compute: $$g''(x)=0.7\ln5\ln2\left(\frac{1}{2^x}-\frac{1}{5^x}\right)$$ Since $x_c<0$, then $g''(x_c)<0$. And that means that $x_c$ provides a local (in fact global) maximum for $g$.  Well, $g(x_c)$ seems ugly to keep in exact form.  It's approximately $-0.14$.  This means $g$ has a negative maximum value and therefore $g(x)$ cannot equal $0$.  (This would not be the case if we had say $0.9$ in place of $0.7$.) Hence $f'(x)$ cannot equal $0$.  And therefore the only solution to the original system posed by the OP is the observed solution with $x=-1$.
(And that leads to $y=-1$ and $\frac{1}{x+y}=-\frac{1}{2}$.)