if $2^x+5^y=2^y+5^x=\frac{7}{10}$ let $x,y$ such
$$2^x+5^y=2^y+5^x=\dfrac{7}{10}$$
prove or disprove $x=y=-1$ is the only solution for the system.
My  try: since
$$2^x-2^y=5^x-5^y$$
But How can prove  or disprove  $x=y$?

Write $x=u-v$, $\>y=u+v$ and put $r:={5\over2}$. Then we have to solve the equations
$$2^{u-v}+5^{u+v}=5^{u-v}+2^{u+v}={7\over10}\ .$$
Dividing the first equation by $2^u$ we obtain
$$(s:=)\qquad 2^{-v} +r^u 5^v = r^u 5^{-v}+ 2^v\ ,$$
or
$$r^u(5^v-5^{-v})=2^v-2^{-v}\ .\tag{1}$$
Equation $(1)$ is obviously fulfilled when $v=0$ and $u$ is arbitrary. This corresponds to an arbitrary choice of $(x,y)$ on the line $x=y$ and leads together with the remaining equation to $x=y=-1$.
But this is not all: For given  $v\ne0$ equation $(1)$ determines a unique $u\in{\mathbb R}$ by means of
$$r^u={2^v-2^{-v}\over 5^v-5^{-v}}\ ,\tag{2}$$
and for this value of $u$ (an even function of $v$) we then get
$$s={1\over2}\bigl(2^v+2^{-v}+r^u(5^v+5^{-v})\bigr)={10^v-10^{-v}\over 5^v-5^{-v}}\ .$$
With the help of $(2)$ it follows that
$$2^u s=\left({2^v-2^{-v}\over 5^v-5^{-v}}\right)^{\!\log 2/\log r}\ {10^v-10^{-v}\over 5^v-5^{-v}}=: f(v)\ .$$

Plotting $f(v)$ one finds that it is minimal at $v=0$ and assumes the (limiting) value $0.756463$ there, which is $>{7\over10}$. It follows that there are no solutions of the original problem with $v\ne0$.
