$5^{x}+2^{y}=2^{x}+5^{y} =\frac{7}{10}$ Work out the values of $\frac{1}{x+y}$

Solution 1:

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 $\frac{1}{x+y}=-\frac{1}{2}$.)

Solution 2:

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 :

1. $\frac{d}{dx}f(x)=0 \Rightarrow x\in \varnothing$ (no real solutions)
2. $\displaystyle\lim_{x \to -\infty} \frac{d}{dx}f(x)=0 $
3. $\displaystyle\lim_{x \to \left(\log_2{\left(\frac{7}{10}\right)}\right)^{-}} \frac{d}{dx}f(x)=-\infty$

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 : $\frac{1}{x+y}=-\frac{1}{2}$

Solution 3:

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, $1/(x+y)=-1/2$

Solution 4:

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.