<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] My solution is quite elementary.
So we consider the equations $$5^x + 2^y = \frac {7}{10} = 2^x +5^y  $$
Note that $5^x + 2^y = \frac {7}{10} ( eq.1 )  $ and $\frac {7}{10} = 2^x +5^y (eq.2) $ are  inverse functions.
By inspection, we can see that  $(-1, -1 )$ is a solution. Now, If we show that this is the only solution then we can clearly state that $$\frac {1}{x+y} = -\frac {1}{2}$$.
Let us now find the domain and the range of the functions. Since the two functions are inverses , the the domain  of the first function is the range of the other function.
Consider only the function $5^x + 2^y = \frac {7}{10}  $.
Solving for $x$ and $y$ , we get $x = log_{5}(\frac {7}{10} - 2^y )$ and $y = log_{2}(\frac {7}{10} - 5^x )$ respectively.
We know that given $y = log x$ ,y is defined if $x> 0$. So, we have the equations $\frac {7}{10} - 2^y > 0 $ and$ \frac {7}{10} - 5^x >0$ 
Solving for x and y,$ y < log_{2}\frac {7}{10} $ and $x < log_{5}\frac {7}{10}$. 
Clearly, the domain and the range are $$\{ x|x < log_{5}\frac {7}{10}\} $$ and $$\{ y|y < log_{2}\frac {7}{10}\}$$.
Therefore, the asymptotes of the first equation  are $x = log_{5}\frac {7}{10}$ and $ y = log_{2}\frac {7}{10}\ $, the asympotes of the second equation are $y = log_{5}\frac {7}{10}$ and $ x = log_{2}\frac {7}{10}\ $, and we know that $ log_{5}\frac {7}{10} <  log_{2}\frac {7}{10}$.  
From this, We can conclude that as the $ x \rightarrow -\infty $ from $-1$. then for each y, $y_{ eq 1} < y_{ eq 2}$ and as $ y \rightarrow -\infty $ from $-1$. then for each x, $ x_{ eq 1} > x_{ eq 2} $. This is clear due to the asymptotes and the idea that they are inverses.
Furthermore, If we show that indeed $ x \rightarrow -\infty $ from $-1$. then, for each y, $y_{ eq 1} < y_{ eq 2}$, then it follows that $ y \rightarrow -\infty $ from $-1$. then for each x, $ x_{ eq 1} > x_{ eq 2} $.
Hence, to show this, notice that $$5^x + 2^y = 5^y +2^x =\frac {7}{10} = \frac {2+5}{10}    =\frac {2}{10}+\frac {5}{10} =\frac {1}{5}+\frac {1}{2} = 5^{-1} + 2^{-1} $$
For the equality to be correct as we make $ x \rightarrow -\infty $ from $-1$, and since $5^x < 2^x$ since $x< -1$, then the decrease in the equation 1 is larger and it requires smaller value of y to be equal to $5^{-1} + 2^{-1}$. 
thank you.
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