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

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

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By inspection, $x = y = -1$ is a solution. Proving that it's the only solution or that other solutions exist is another matter though... – Sp3000 Nov 20 '11 at 8:35
plot($\log_{5}(\frac{7}{10}-2^x)-\log_{2}(\frac{7}{10}-5^x)$),link – pedja Nov 20 '11 at 10:02
It's easy to see that $x=y=-1$ is the only solution where $x=y$. By itself, the equation $5^x+2^y = 2^x+5^y$ is equivalent to $5^x-2^x = 5^y-2^y$. The function $f(x)=5^x-2^x$ is negative for $x<0$ and positive thereafter; also, $f(x)$ is decreasing to the left of $x_0=-\log_{5/2} (\log_2 5)$ and increasing thereafter. So for each $x<0$ ($x\ne x_0$) there is exactly one $y\ne x$ such that $5^x-2^x = 5^y-2^y$. - Having said all that, I don't immediately see how to proceed from here. – Greg Martin Nov 20 '11 at 11:22
We can see from @Greg's comment that there would be further solutions if we replace $7/10$ by a number between $2^\alpha+5^\alpha\approx0.7565$ and $1$, where $\alpha=-(\log\log5-\log\log2)/(\log5-\log2)\approx-0.9194$. Thus the proof that $x=y=-1$ is the only solution must somehow use the fact that $7/10$ is outside this interval. I don't see how any of the answers given so far do that. – joriki Dec 6 '11 at 8:35
Given that there is a bifurcation in the number of solutions as the parameter $A$ moves from $0.7$ through $0.9$, and that the exact value of $A$ at the bifurcation is a horribly messy thing that can be computed using my answer (and subsequent comment) below, I suspect that a calculus-free proof of a unique solution for $(x,y)$ is not possible. I'd love to see one though. – alex.jordan Dec 8 '11 at 4:26

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

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I like your answer better. – Kou Dec 8 '11 at 3:56
Replacing the $0.7$ with a parameter $A$, we can solve for the value of $A$ where $g(x_c)=0$. I think the exact form is too messy for the comment section, but this critical value of $A$ is about $0.756463$. This is in agreement with @joriki's measurement (in the comments) of where the parameter $A$ begins to admit more that one solution. – alex.jordan Dec 8 '11 at 4:04

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

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Please see my comment under the question -- where does your proof make use of the fact that $7/10$ is outside the interval where there would be additional solutions? – joriki Dec 6 '11 at 8:38
@joriki,honestly, I don't understand what it means to "replace $\frac{7}{10}$ by number between..." – pedja Dec 6 '11 at 9:32
If we ask the same question, but with, say, $9/10$ instead of $7/10$, then in addition to the solution with $x=y$, there will be another solution as indicated in Greg's comment. I was asking where in your proof you make specific use of $7/10$ such that the argument wouldn't work with $9/10$ instead. I see now that your result 1) that the derivative of $f$ doesn't vanish is specific to $7/10$. As you pointed out at the beginning of your answer, you haven't justified this result beyond a calculation by Wolfram|Alpha. – joriki Dec 6 '11 at 10:11
I can't find a way to show your item 1 by hand. Surely it would rely on the value $7/10$, as @joriki is saying. I'm not sure what WolframAlpha is doing, but it's not trivial; I can't verify that what it is doing is rigorously correct. – alex.jordan Dec 8 '11 at 1:14
@alex.jordan,thanks for edit, great improvement.... :-) – pedja Dec 8 '11 at 5:57

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

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Please see my comment under the question -- where does your proof make use of the fact that $7/10$ is outside the interval where there would be additional solutions? – joriki Dec 6 '11 at 8:38

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.

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Please see my comment under the question -- where does your proof make use of the fact that $7/10$ is outside the interval where there would be additional solutions? – joriki Dec 6 '11 at 8:37

This is not yet a complete answer, but another approach, and avoids calculus.
I'm trying this because that sort of problem occurs often and I would like to have a somehow general path for a solution.

The idea is, that the two equations $\small 5^x + 2^y = 0.7$ and $\small 5^y+2^x=0.7$ define both bounds for x and y . If we do an initial estimate for upper and lower bounds, then we see, that by the estimate of one bound determines another one and that this might be then some better estimate. After iterations of estimations and computations these bounds may then narrow to a unique solution. Actually they approach x=y=-1 for this case where the constant in the equation is 7/10.

This approach does not yet satisfy joriki's call for an analytic argument for the exclusivity of that solution with constant C=7/10 (at least for the constant 9/10 it correctly does not converge to a single solution but to two values for x and y ).
Also it is not yet a proof but merely an idea from where a proof might be derived.

We may first rewrite $\small 5^x + 2^y = 0.7$ as deviation from the known solution to get $\small 5^{-1+a} + 2^{-1+b} = 0.7 =C$ and the other equivalently or even better (and the exponents simplified to a and b to avoid the additional -1): $$\small \begin{array} {lll} A) \qquad {1\over 5^a }+ {1\over 2^b} = C \\ B) \qquad {1\over 5^b }+ {1\over 2^a} = C \end{array}$$

Obviously an upper limit for a and b is infinity and if we set one of the variables to infinity the other can be determined by a formula $$\small \begin{array} {lll} A) & {1\over 5^a }+ {1\over 2^\infty} = C & \to & a_{l,A}= \log_{1/5}(C) \\ A) & {1\over 5^\infty }+ {1\over 2^b} = C & \to & b_{l,A}= \log_{1/2}(C) \\ \end{array}$$ If we assume for a the "infinity" as an upper bound (second row), then for b the value $\small b_{l,A}$ is a lower bound. And equvalently for a is $\small a_{l,A}$ a lower bound if we assume "infinity" as upper bound for b.

If we use equation B) we get other values, write them as $\small a_{l,B}$ and $\small b_{l,B}$.

But because both equations must be satisfied simultaneously, the lower bound of a must be the greater one of the two results, and the same occurs for b, so we get $$\small a_l = max(a_{l,A}, a_{l,B}) \qquad \qquad b_l = max(b_{l,A}, b_{l,B})$$

But now, because in one of A) or in B) the exponent a is bigger than possible (by the lower bound of this equation when b is "infinity", the exponent at b must decrease from "infinity" down to a new upper bound. So this defines then a lower upper-bound for b and analoguously for a by the other equation.

So by one turn we have now narrowed our lower and upper bounds for both variables and can begin to iterate.

Heuristically we seem to get convergence for C=7/10, and for C=9/10 we get convergence to two different values.

The idea, how to make a proof out of this is still somehow vague and I do not know, whether I can manage to explicate a formula for this interwoven functional iteration.

Some Pari/GP-Code

[l5=log(1/5),l2=log(1/2)]

[C = 7/10, INF =999 ]
a_A(b)=local(a);a=C-if(b>=INF,0,1/2^b);a=log(a)/l5
b_A(a)=local(b);b=C-if(a>=INF,0,1/5^a);b=log(b)/l2
a_B(b)=local(a);a=C-if(b>=INF,0,1/5^b);a=log(a)/l2
b_B(a)=local(b);b=C-if(a>=INF,0,1/2^a);b=log(b)/l5

a_Min(b)=min(a_A(b),a_B(b))
b_Bin(a)=min(b_A(a),b_B(a))
a_Max(b)=max(a_A(b),a_B(b))
b_Max(a)=max(b_A(a),b_B(a))

a_up=b_up=INF
\\ iterate to convergence
[a_lo=a_Max(b_up),b_lo=b_Max(a_up) ; a_up=a_Min(b_lo),b_up=b_Min(a_lo)]


It is interesting, that a and b have the same upper and the same lower bounds after the first iteration: I think this should help for some line of the proof.

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As @ken-dee and others have pointed out, the inverse graphs $$F=\{(x,y)|5^x+2^y=c\},$$ $$G=\{(x,y)|2^x+5^y=c\}$$ define inverse functions $f(x)=log_2(c-5^x),\quad g(x)=log_5(c-2^x)$ for $c=\frac{7}{10}$, where $f:[-\infty,log_{5}c)\to[-\infty,log_{2}c]$ and $g$ goes in the other direction, each having horizontal and vertical asymptotes at the upper endpoints of their respective domains and ranges, as can be seen on the graph below. Since $f$ and $g$ are inverse functions, they are symmetric about the line $y=x$, and so must meet along the line $y=x$ wherever they intersect, for example at $(-1,-1)$. But both are decreasing functions as shown below, so there can be no other real solutions (as can be seen from the graph below), for this would require a positive average slope of $1$ between the other, hypothetical solution $(x,x)$ and the known one, $(-1,-1)$. Therefore, as already argued, $\frac{1}{x+y}=-\frac{1}{2}$ has only one value on $F \cap G=\{(-1,-1)\}$.

As stated, the functions $f$ and (its inverse) $g$ are everywhere decreasing: $$f^\prime(x) =\frac{d}{dx}log_2(c-5^x) =\frac{(c-5^x)^{-1}}{\ln 2}\frac{d}{dx}(c-5^x) \quad\implies$$ $$f^\prime(x) =(c-5^x)^{-1}\cdot -5^x \cdot\frac{\ln 5}{\ln 2} =-(\log_{2}{5})(c\cdot5^{-x}-1)^{-1} <0$$ since $$x\in[-\infty,log_{5}c) \quad\implies\quad 5^x<c \quad\implies\quad c\cdot5^{-x}>1 \quad\implies\quad c\cdot5^{-x}-1>0$$ where we have used the facts that $\log_ax=\frac{\ln x}{\ln a}$ & $a^x=e^{x\ln a}$ for $a,x>0$, that $(e^u)^\prime=e^u\cdot u^\prime$, that $(\ln u)^\prime=\frac{u^\prime}{u}$, and lastly, that $g^\prime(y)=\frac{1}{f^\prime(x)}$ at any point $(x,y)$ on the graph of $f$ for inverse functions $f$ & $g$ (inverse functions have reciprocal slopes at corresponding points), so that $g$ is likewise decreasing.

Here is some sage code and a plot to illustrate the numerical situation. The first function $f$ and its asymptotes are shown in blue, and the second function $g$ and its asymptotes are in green. The lesser asymptotes are dashed, the greater are dotted.

x,y = var('x,y'); c = 7/10
f(x,y) = 5^x + 2^y - c; xf = log(c)/log(5)
g(x,y) = 2^x + 5^y - c; yf = log(c)/log(2)
[xf.n(digits=6), yf.n(digits=6)] # [-0.221615, -0.514573]
G = Graphics(); G.set_aspect_ratio(1)
G += implicit_plot(f    , (x,-3,0), (y,-3,0), color="blue",  legend_label="5^x + 2^y - 7/10")
G += implicit_plot(g    , (x,-3,0), (y,-3,0), color="green", legend_label="2^x + 5^y - 7/10")
G += implicit_plot(x==xf, (x,-3,0), (y,-3,0), color="blue",  linestyle='dotted')
G += implicit_plot(y==yf, (x,-3,0), (y,-3,0), color="blue",  linestyle='dashed')
G += implicit_plot(x==yf, (x,-3,0), (y,-3,0), color="green", linestyle='dashed')
G += implicit_plot(y==xf, (x,-3,0), (y,-3,0), color="green", linestyle='dotted')
G += implicit_plot(y==x , (x,-3,0), (y,-3,0), color="gray",  linestyle='dashdot')
G.show()


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When $c=0.8$, your curves cross three times. – Henry Dec 8 '11 at 1:50