Why are there $3$ solutions to $x^a = a^x$?

Question

Consider the equation, $x^a = a^x$.

If a is of the form $1.6 + 0.4k$, where $k$ is a natural number, then the equation has $3$ solutions. I checked them until $k = 1000$, but was unable to find a reason for this. Is there any plausible explanation for this? If yes, what is it? if not, why?

The possible duplicate question doesn't explain why there are 3 solutions for this particular form!

Thanks, John.

I am understanding that for each integer $k$, you are fixing the value of $a$ and solving for $x$, am I correct? — Krishnarjun, Nov 20 '19 at 18:30
@CristhianPardo, There are 3 real solutions when k = 1, (x = 2,4,and a negative value). — John Paul, Nov 20 '19 at 18:49
@Krishnarjun, Yes I got negative solutions as well, I tried rough sketching for few integral values of (a) and later graphed them on Desmos. — John Paul, Nov 20 '19 at 18:49
Does there exist a negative solution for non integral values of $a$? — Krishnarjun, Nov 20 '19 at 19:22
@Krishnarjun, Yes there are 3 solutions for non integral values of a either, you can check it for yourself, here . : https://www.desmos.com/calculator/fhkmhptwxw — John Paul, Nov 20 '19 at 19:28

score 1 · Answer 1 · 2019-11-20T20:22:34.890

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The two positive solutions

On $x>0$, the value of $\frac{\ln x}{x}$ is monotonic increasing from $-\infty$ to $\frac{1}{e}$ and then monotonic decreasing, tending to $0$.

Therefore, for $a>1$ and $a\ne e$, the horizontal line $y=\frac{\ln a}{a}$ will be cut precisely twice by $\frac{\ln x}{x}$.

This will be obvious to you if you sketch the graph.

Negative solutions?

Instead of taking $x$ to the power $a=1.6 +0.4k$ you could consider not $x^a$ but $(x^2)^{0.8+0.2k}.$

You now have a fractional power of a positive number and so you can happily equate this expression with $a^x$. But please note that this is not the same as solving your original equation.

(This is, I believe, why you were convinced that you had solutions which others were saying did not exist.)

edited Nov 20 '19 at 20:22

answered Nov 20 '19 at 19:13

I don't get why y = ln(a)/a cuts ln(x)/x 2 times, can you elaborate it please?, what I am thinking is ln(a)/a is a point and it shall cut the graph only once. And for the negative solution, for every x,a > 1, there will be definitely a negative solution. (Just by graphing x^a and a^x) – John Paul Nov 20 '19 at 19:30
Think of ln(a)/a as a number e.g. 2. Then we are finding where ln(x)/x cuts the line y=2. – Nov 20 '19 at 19:33
Ah, Got it, Thanks !. But, according to you, wouldn't every value of a has 2 cuts? – John Paul Nov 20 '19 at 19:37
P.S. Sort out your understanding of the positive solutions first and then, if necessary, I'll explain what the issue is re. negative solutions. – Nov 20 '19 at 19:38
For a=e the curve just touches the line. For $a\le 1$ the line is below the x-axis and is only cut once. Of course all the values you are considering are not affected by this. – Nov 20 '19 at 19:40
So what you are saying is, as long as a > 1 and a != e, we have 2 pos. solutions for x^a = a^x, right? – John Paul Nov 20 '19 at 19:42
Yes that is correct. – Nov 20 '19 at 19:43
Ok so, all we have to see is why a = 1.6+0.4*k adds a negative solution,right? – John Paul Nov 20 '19 at 19:46
Yes. However, there is a relatively complicated issue with fractional powers of negative numbers. For example, while (-1)^n is fine for n an integer it is not real for n=1/2. – Nov 20 '19 at 19:52
But a isn't negative fractional number, it is positive, so how does this complicate? – John Paul Nov 20 '19 at 19:58
You are considering x to the power a where x is negative and some of the a are not integers. – Nov 20 '19 at 20:00
Got it, Thanks for the clarification ! , So, there is no plausible explanation for negative solutions? – John Paul Nov 20 '19 at 20:04
I will add a way of getting negative solutions to my solution - it will take a few minutes. – Nov 20 '19 at 20:06
You haven't proved why there is a negative solution whenever a is 1.6+0.4*k, and yes, a negative solution does exist ! – John Paul Nov 21 '19 at 14:13

score 0 · Answer 2 · answered Nov 20 '19 at 18:50

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Taking the logarithm on both sides we get $$\frac{\ln(x)}{x}=\frac{\ln(a)}{a}$$ Now you can discover how many times cut $$\frac{\ln(a)}{a}$$ the graph of $$\frac{\ln(x)}{x}$$

answered Nov 20 '19 at 18:50

Dr. Sonnhard Graubner

95,283

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But ln(a)/a is a constant, so wouldn't it cut the graph at a single point? – John Paul Nov 20 '19 at 18:53
You must consider the function $$f(x)=\frac{\ln(x)}{x}$$ by calculus – Dr. Sonnhard Graubner Nov 20 '19 at 18:55
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Doesn't ln(a)/a being a constant cut the graph of f(x) at a single point? – John Paul Nov 20 '19 at 18:59
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$\frac{\ln(x)}{x}$ can cut $f(x)$ two times – Dr. Sonnhard Graubner Nov 20 '19 at 19:00
how can a graph cut itself 2 times? what do you mean? Could you please clarify up a bit?Thanks,John. – John Paul Nov 20 '19 at 19:03
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Sorry i mean $$\frac{\ln(a)}{a}$$ – Dr. Sonnhard Graubner Nov 20 '19 at 19:06
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If $x<0$, then $\log x$ is not well defined, therefore this argument does not give you the negative solutions, which are claimed to exist – Krishnarjun Nov 20 '19 at 19:07
@Krishnarjun, Yes, that's true. – John Paul Nov 20 '19 at 19:10

score 0 · Answer 3 · answered Nov 20 '19 at 19:45

This is a partial answer.

Let us say that $a\neq 1$ and also $x\neq 1$, then we are interested in finding $x$ such that $$\frac{a}{\log a}=\frac{x}{\log x}.$$ If you consider the given domain, (by plotting a graph using a computer), we can see that there are precisely two different values for $x$ which satisfy this equation. This takes care of the two positive solutions. The third solution, if it exists should therefore be negative.

Fix $a>0$. Suppose there is a negative solution, call it $-y$ where $y>0$. Therefore we have that $$(-y)^a=\frac{1}{a^y}\Rightarrow (-1)^a(y)^a=\frac{1}{a^y}\Rightarrow \left(\cos(a\pi)+i\sin(a\pi)\right)(y)^a=\frac{1}{a^y}.$$

Since the right hand side is a real number (since $a>0$ and $y>0$), we should have that $\sin(a\pi)=0$, that is $a\in\mathbb{Z}$. Therefore whenever $a$ is non integral, there does not exist three solutions. Furthermore when $a$ is odd, we will have $$- y^a = a^{-y}$$ which is not possible because the LHS is a negative number and the RHS is a positive number (Again since $a>0$ and $y>0$). Therefore for a negative solution to exist, $a$ has to be even. In this case we are interested in solving, $y^a = a^{-y}$, where once again the $\log$ trick can be used.

In order to explain the third negative root in the case of non integral values of $a$, I think it is necessary to consider $y^a$ and $a^y$ as complex numbers. I observed that for $a=3$, there are only two solutions. (Also your formula for $a$ will only give even integral values for $a$. It will not give odd integral values) — Krishnarjun, Nov 20 '19 at 19:48
"Therefore whenever a is non integral, there does not exist three solutions". check by putting a = 1.6. And, I am just asking "Why are there 3 solutions whenever a is 1.6+0.4*k" , it doesn't matter if I don't include odd integral values of a, all I am asking is about this form. I don't know whether you have done wrong, but the conclusion is definitely not right. — John Paul, Nov 20 '19 at 20:02

Why are there $3$ solutions to $x^a = a^x$?

3 Answers3