$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ 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 $\frac{1}{x+y}$
 A: 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 $\frac{1}{x+y}=-\frac{1}{2}$.)
A: 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 : $\frac{1}{x+y}=-\frac{1}{2}$
A: 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,
$1/(x+y)=-1/2$
A: 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.
A: This is not yet a complete answer, but another approach, and avoids calculus.
I'm trying this because that sort of problem occurs often and I would like
to have a somehow general path for a solution.
The idea is, that the two equations 
$\small 5^x + 2^y = 0.7 $ and $\small 5^y+2^x=0.7 $ define both bounds 
for x and y . If we do an initial estimate for upper and lower bounds, 
then we see, that by the estimate of one bound determines another one
and that this might be then some better estimate. After iterations of
estimations and computations these bounds may then narrow to a unique
solution. Actually they approach x=y=-1 for this case where the constant 
in the equation is 7/10.    
This approach does not yet satisfy joriki's call for an analytic argument 
for the exclusivity of that solution with constant C=7/10 (at least for 
the constant 9/10 it correctly does not converge to a single solution 
but to two values for x and y ).
Also it is not yet a proof but merely an idea from where a proof might
be derived.
We may first rewrite $\small 5^x + 2^y = 0.7 $ as deviation from the known
solution to get $\small 5^{-1+a} + 2^{-1+b} = 0.7 =C $ and the other equivalently
or even better (and the exponents simplified to a and b to avoid the 
additional -1):
$$ \small \begin{array} {lll} 
 A) \qquad {1\over 5^a }+ {1\over 2^b} = C \\
 B) \qquad {1\over 5^b }+ {1\over 2^a} = C
 \end{array} $$ 
Obviously an upper limit for a and b is infinity and if we set
one of the variables to infinity the other can be determined by a formula
$$ \small \begin{array} {lll} 
 A) & {1\over 5^a }+ {1\over 2^\infty} = C & \to & a_{l,A}= \log_{1/5}(C) \\
 A) & {1\over 5^\infty }+ {1\over 2^b} = C & \to & b_{l,A}= \log_{1/2}(C) \\
 \end{array} $$ 
If we assume for a the "infinity" as an upper bound (second row), then 
for b  the value $\small b_{l,A} $ is a lower bound. And equvalently 
for a is $\small a_{l,A} $ a lower bound if we assume "infinity" as
upper bound for b.      
If we use equation B) we get other values, write them as $\small a_{l,B} $
and $\small b_{l,B} $.     
But because both equations must be satisfied simultaneously, the lower
bound of a must be the greater one of the two results, and the same occurs 
for b, so we get 
$$\small a_l = max(a_{l,A}, a_{l,B}) \qquad \qquad b_l = max(b_{l,A}, b_{l,B}) $$
But now, because in one of A) or in B) the exponent a 
is bigger than possible (by the lower bound of this equation when b is "infinity",
the exponent at b must decrease from "infinity" down to a new upper bound. 
So this defines then a lower upper-bound for b and analoguously for a 
by the other equation.      
So by one turn we have now narrowed our lower and upper bounds for
both variables and can begin to iterate.
Heuristically we seem to get convergence for C=7/10, and for C=9/10
we get convergence to two different values.    
The idea, how to make a proof out of this is still somehow vague
and I do not know, whether I can manage to explicate a formula
for this interwoven functional iteration.
Some Pari/GP-Code
[l5=log(1/5),l2=log(1/2)]

[C = 7/10, INF =999 ]
a_A(b)=local(a);a=C-if(b>=INF,0,1/2^b);a=log(a)/l5
b_A(a)=local(b);b=C-if(a>=INF,0,1/5^a);b=log(b)/l2
a_B(b)=local(a);a=C-if(b>=INF,0,1/5^b);a=log(a)/l2
b_B(a)=local(b);b=C-if(a>=INF,0,1/2^a);b=log(b)/l5

a_Min(b)=min(a_A(b),a_B(b))
b_Bin(a)=min(b_A(a),b_B(a))
a_Max(b)=max(a_A(b),a_B(b))
b_Max(a)=max(b_A(a),b_B(a))

a_up=b_up=INF
\\ iterate to convergence
[a_lo=a_Max(b_up),b_lo=b_Max(a_up) ; a_up=a_Min(b_lo),b_up=b_Min(a_lo)]

It is interesting, that a and b have the same upper and the
same lower bounds after the first iteration: I think this should help for 
some line of the proof.
A: As @ken-dee and others have pointed out, the inverse graphs
$$
F=\{(x,y)|5^x+2^y=c\},
$$
$$
G=\{(x,y)|2^x+5^y=c\}
$$
define inverse functions $f(x)=log_2(c-5^x),\quad g(x)=log_5(c-2^x)$
for $c=\frac{7}{10}$,
where $f:[-\infty,log_{5}c)\to[-\infty,log_{2}c]$ and $g$ goes in the other direction, each having horizontal and vertical asymptotes at the upper endpoints of their respective domains and ranges, as can be seen on the graph below.
Since $f$ and $g$ are inverse functions, they are symmetric about the line $y=x$, and so must meet along the line $y=x$ wherever they intersect, for example at $(-1,-1)$. But both are decreasing functions as shown below, so there can be no other real solutions (as can be seen from the graph below), for this would require a positive average slope of $1$ between the other, hypothetical solution $(x,x)$ and the known one, $(-1,-1)$. Therefore, as already argued, $\frac{1}{x+y}=-\frac{1}{2}$ has only one value on $F \cap G=\{(-1,-1)\}$.
As stated, the functions $f$ and (its inverse) $g$ are everywhere decreasing:
$$
f^\prime(x)
=\frac{d}{dx}log_2(c-5^x)
=\frac{(c-5^x)^{-1}}{\ln 2}\frac{d}{dx}(c-5^x)
\quad\implies
$$
$$
f^\prime(x)
=(c-5^x)^{-1}\cdot -5^x \cdot\frac{\ln 5}{\ln 2}
=-(\log_{2}{5})(c\cdot5^{-x}-1)^{-1}
<0
$$
since
$$
x\in[-\infty,log_{5}c)
\quad\implies\quad 5^x<c
\quad\implies\quad c\cdot5^{-x}>1
\quad\implies\quad c\cdot5^{-x}-1>0
$$
where we have used the facts that 
$\log_ax=\frac{\ln x}{\ln a}$ &
$a^x=e^{x\ln a}$ for
$a,x>0$, that
$(e^u)^\prime=e^u\cdot u^\prime$, that
$(\ln u)^\prime=\frac{u^\prime}{u}$, and lastly, that
$g^\prime(y)=\frac{1}{f^\prime(x)}$ at any point $(x,y)$ on the graph of $f$ for inverse functions $f$ & $g$ (inverse functions have reciprocal slopes at corresponding points), so that $g$ is likewise decreasing.
Here is some sage code and a plot to illustrate the numerical situation. The first function $f$ and its asymptotes are shown in blue, and the second function $g$ and its asymptotes are in green. The lesser asymptotes are dashed, the greater are dotted. 
x,y = var('x,y'); c = 7/10
f(x,y) = 5^x + 2^y - c; xf = log(c)/log(5)
g(x,y) = 2^x + 5^y - c; yf = log(c)/log(2)
[xf.n(digits=6), yf.n(digits=6)] # [-0.221615, -0.514573]
G = Graphics(); G.set_aspect_ratio(1)
G += implicit_plot(f    , (x,-3,0), (y,-3,0), color="blue",  legend_label="5^x + 2^y - 7/10")
G += implicit_plot(g    , (x,-3,0), (y,-3,0), color="green", legend_label="2^x + 5^y - 7/10")
G += implicit_plot(x==xf, (x,-3,0), (y,-3,0), color="blue",  linestyle='dotted')
G += implicit_plot(y==yf, (x,-3,0), (y,-3,0), color="blue",  linestyle='dashed')
G += implicit_plot(x==yf, (x,-3,0), (y,-3,0), color="green", linestyle='dashed')
G += implicit_plot(y==xf, (x,-3,0), (y,-3,0), color="green", linestyle='dotted')
G += implicit_plot(y==x , (x,-3,0), (y,-3,0), color="gray",  linestyle='dashdot')
G.show()


