<s>[INST] $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$ $5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$
Work out the values of $\frac{1}{x+y}$
 [/INST] Remark : This answer is based on WolframAlpha calculations .
As Sp300 pointed out in his comment by inspection we can see that $x=-1$ is a solution. I will try to show that this is the only solution.
So :
$$2^y=\frac{7}{10}-5^x \Rightarrow y= \log_2{\left(\frac{7}{10}-5^x\right)}$$
$$5^y=\frac{7}{10}-2^x \Rightarrow y= \log_5{\left(\frac{7}{10}-2^x\right)}$$
Therefore we may conclude that :
$$\log_2{\left(\frac{7}{10}-5^x\right)}=\log_5{\left(\frac{7}{10}-2^x\right)}$$
Let's define $f(x)$ as :
$$f(x)=\log_5{\left(\frac{7}{10}-2^x\right)}-\log_2{\left(\frac{7}{10}-5^x\right)}$$
Note that $f(x)$ is defined for $x\in \left(-\infty,\log_2{\left(\frac{7}{10}\right)}\right)$
WolframAlpha gives following results :


*

*$\frac{d}{dx}f(x)=0 \Rightarrow x\in \varnothing$ (no real
solutions)

*$\displaystyle\lim_{x \to -\infty} \frac{d}{dx}f(x)=0 $

*$\displaystyle\lim_{x \to
    \left(\log_2{\left(\frac{7}{10}\right)}\right)^{-}}
    \frac{d}{dx}f(x)=-\infty$


From this above we may conclude that $f'(x) < 0$ for all $x\in\left(-\infty,\log_2{\left(\frac{7}{10}\right)}\right)$ , so $f(x)$ is decreasing on interval $x\in \left(-\infty,\log_2{\left(\frac{7}{10}\right)}\right)$ therefore , $x=-1$ is the only solution .
From the equation $y= \log_2{\left(\frac{7}{10}-5^x\right)}$ we can obtain value of $y$ , so $y=-1$
Finally : $\frac{1}{x+y}=-\frac{1}{2}$
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