$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ 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 $\frac{1}{x+y}$

Initial Solution:
$5^x+2^y=7/10$
$5^x+2^y=(5+2)/(5*2)=(1/2+1/5)=5^{-1}+2^{-1}$ ------ (1)
$5^y+2^x=(5+2)/(5*2)=(1/5+1/2)=5^{-1}+2^{-1}$  ------ (2)
By examining (1) and (2) we get the solution:
$x=y=-1$
Now we consider the relation:
$5^x+2^y=k$ --------(1)
where k is a constant.
$\frac{dy}{dx}=-\frac{5^x}{2^y}log_{2} 5<0$ for all values of x and y. So y is a decreasing function of x.
We consider another relation:
$5^y+2^x=k$ --------------(2)
where k is a constant.
$\frac{dy}{dx}=-\frac{2^x}{5^y}\frac{1}{log_2 5}<0$
Again , the function considered is a decreasing one.
But the slope of the first function[expressed by relation (1)] is always greater in magnitude than the slope of the second one[expressed by relation (2)] For a particular point on the x-axis the curve of the first function has a steeper slope than the second one. So their curves  can intersect at only one point. 
Incidentally to show the greater steepness/slope of the first curve we may write,
$\frac{5^x}{2^y}log_{2} 5-\frac{2^x}{5^y}\frac{1}{log_{2} 5}$
$=\frac{5^{x+y}(log_{2} 5)^2-2^{x+y}}{10^y log_{2} 5}>0$
Therefore,
$1/(x+y)=-1/2$
