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  $x=y$?
 A: 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 $I(x)D(x) < I(-1.65)\lim_{x\to-\infty} D(x)$.)
A: 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 $v\ne0$.
A: Here is my idea of proving it:


*

*For $x=y=-1$, you can verify that $2^{-1}+5^{-1}=0.7$

*In order to refute any other possible solution where $x=y$:


*

*You can prove that $f(x)=2^x+5^x$ is monotonously increasing

*Do it by showing that $f'(x)=2^x\ln2+5^x\ln5$ is always positive


*In order to refute any other possible solution where $x \neq y$:


*

*You have the following two equations:


*

*$5^y=0.7-2^x$

*$2^y=0.7-5^x$


*Write down each equation as a simple function:


*

*$y=\log_5(0.7-2^x)$

*$y=\log_2(0.7-5^x)$


*Prove that if $x \neq -1$, then the functions are not equal:


*

*Prove that if $x>-1$, then $\log_2(0.7-5^x)>\log_5(0.7-2^x)$

*Prove that if $x<-1$, then $\log_2(0.7-5^x)<\log_5(0.7-2^x)$




Here is the graph of both functions, intersecting at $(-1,-1)$:

