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  $x=y$?

If $(x,y)$ is a solution, then we have both $2^y=\frac7{10}-5^x$ and
$$
2^y = (5^y)^{\log2/\log5} = \big(\tfrac7{10}-2^x\big)^{\log2/\log5}.
$$
So define $f(x)=\frac7{10}-5^x$ and $g(x)=\big(\tfrac7{10}-2^x\big)^{\log2/\log5}$. We want to show that the only place $f$ and $g$ are equal is at $x=-1$; it suffices to show that $f'(x)<g'(x)$ everywhere (to the right of $x=\log_2(0.7)$), or equivalently that $f'(x)/g'(x)<1$.
A calculation shows that $f'(x)/g'(x) = I(x)D(x)$, where
$$
I(x) = \bigg(\frac{\log5}{\log2}\bigg)^2\left(\frac{5}{2}\right)^x \quad\text{and}\quad D(x) = \left(\frac{7}{10}-2^x\right)^{1-\log2/\log5}
$$
are increasing and decreasing functions, respectively. One can thus show that $I(x)D(x) < 1$ on an interval $[a,b]$ by showing that $I(a)D(b)<1$. In this way, one can show separately on each of the intervals
$$
(-\infty,-1.65],\, [-1.65,-1.25],\, [-1.25,-1.05],\, [-1.05,-0.9],\, [-0.9,-0.75],\, [-0.75,\log_2(0.7)] 
$$
that $I(x)D(x) < 1$. (On the leftmost interval, use $I(x)D(x) < I(-1.65)\lim_{x\to-\infty} D(x)$.)
