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

Solution 1:

Here is my idea of proving it:

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

2. 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

3. 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 $(-1,-1)$:

Solution 2:

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 $I(x)D(x) < I(-1.65)\lim_{x\to-\infty} D(x)$.)

Solution 3:

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 $v\ne0$.

Solution 4:

Here's $99\%$ of a proof. Unfortunately I'm stuck on the final, crucial $1\%$.

Edit: Having finally understood Greg Martin's answer, I now see how to get unstuck on the crucial $1\%$. I'm inserting the rest of the proof at the pertinent point, but leaving the rest alone. End edit

Note that $2^x+5^y=2^y+5^x$ implies $5^y-2^y=5^x-2^x$, so let's start by considering the "helper" function $h(x)=5^x-2^x$ on $(-\infty,0)$, which is where any solutions to the OP's problem must lie (since ${7\over10}\lt1$). It's easy to see that $h(x)$ is always negative for $x\lt0$, that $h(0)=0$, and that $\lim_{x\to-\infty}h(x)=0$. Furthermore, $h$ has a single minimum:

$$h'(x)=(\ln5)5^x-(\ln2)2^x=0\implies x={\ln\ln5-\ln\ln2\over\ln5-\ln2}\approx-.91936$$

For convenience, let's call this value $c$.

The upshot of this is that the assumption $y\not=x$ for the equation $2^x+5^y=2^y+5^x$ implicitly defines a function $y(x)$ with $(c,0)$ as its domain and $(-\infty,c)$ as its range.

Consider now the function $f(x)=2^x+5^{y(x)}$ on the interval $(c,0)$. The question is, is ${7\over10}$ in the range of this function? Noting that $\lim_{x\to c}y(x)=c$ and $\lim_{x\to0}y(x)=-\infty$, we have $f(c)=2^c+5^c\approx.75646$ and $f(0)=1$. Since ${7\over10}\lt.75646$, we will be finished if we can show that $f(x)=2^x+5^{y(x)}$ doesn't decrease too much from its value of $.75646$ at $x=c$ before it increases to its value of $1$ at $x=0$.

Insertion: $f(x)$ is the sum of the increasing function $2^x$ and the decreasing function $5^{y(x)}$. Much as in Greg Martin's answer, we can show that $f(x)\gt.7$ on an interval $[a,b]$ if we can show $2^a+5^{y(b)}\gt.7$. Some careful calculation with the helper function $h(x)=5^x-2^x$ shows

$$\begin{align} y(-.85)&\gt-1\\ y(-.75)&\gt-1.15\\ y(-.6)&\gt-1.37\\ y(-.5)&\gt-1.6 \end{align}$$

from which we have

$$\begin{align} 2^c+5^{y(-.85)}&\gt 2^{-.91936}+5^{-1}\approx.7287\\ 2^{-.85}+5^{y(-.75)}&\gt 2^{-.85}+5^{-1.15}\approx.71189\\ 2^{-.75}+5^{y(-.6)}&\gt 2^{-.75}+5^{-1.37}\approx.70486\\ 2^{-.6}+5^{y(-.5)}&\gt 2^{-.6}+5^{-1.6}\approx.7359\\ 2^{-.5}+5^{y(0)}&\gt2^{-.5}+0\approx.707 \end{align}$$

which is all we need to conclude that $f(x)\gt.7$ on $[c,0]$. End insertion

In fact it seems likely that $f(x)$ doesn't decrease at all on $(c,0)$, but that's the $1\%$ I'm stuck on. The most I can say is that $2^x+5^y=2^y+5^x$ implies

$$y'={(\ln5)5^x-(\ln2)2^x\over(\ln5)5^y-(\ln2)2^y}$$

and hence

$$\begin{align} f'(x)&=(\ln2)2^x+(\ln5)y'5^y\\ &={(\ln5)^25^{x+y}-(\ln2)^22^{x+y}\over(\ln5)5^y-(\ln2)2^y}\\ &=\left((\ln5)5^{(x+y)/2}+(\ln2)2^{(x+y)/2}\right){(\ln5)5^{(x+y)/2}-(\ln2)2^{(x+y)/2}\over(\ln5)5^y-(\ln2)2^y} \end{align}$$

In this last expression, the factor in front is clearly positive. As for the rest of it the denominator is negative since $y=y(x)\lt c$. We would also have a negative numerator (and hence a strictly increasing function $f$) if we could show that $(x+y(x))/2\lt c$ for all $x\in(c,0)$. But again, I'm stuck. I thought I'd post things anyway, in case someone can fill in the gaps.