The equation $a^x=x$ for $a>1$.

Question

The following problem arises when I was sketching the inverse of $f(x)=a^x$ graphically for $a>1$:

1) Is there an $a>1$ such that the equation $a^x=x$ has a unique solution $x\in\mathbb{R}$?

2) If so, then how do we find such $a$ explicitly if possible?

The answer to the first question seems to be yes, I tried solving the equation $a^x=x$ for $a=1.4$ and $a=1.5$, which has two and no solutions respectively; but I personally would love to see a proof without geometry argument.

Answer 1

Let $g(x)=a^x-x$. Then

$$g'(x)=(\ln a)a^x-1$$

Assume $(\exists \,\,a>1,x_0>0)\,\,g(x_0)=g'(x_0)=0$. Then using $g'(x_0)=0$,

$$(\ln a)a^{x_0}=1$$ $\implies$ $$a^{x_0}=e^{(\ln a){x_0}}=\frac{1}{\ln a}$$ $\implies$ $$(\ln a){x_0} =\ln\left(\frac{1}{\ln a}\right)$$ $\implies$ $${x_0} =\frac{\ln\left(\frac{1}{\ln a}\right)}{\ln a}$$

Then

$$g(x_0)=a^{\left(\frac{\ln\left(\frac{1}{\ln a}\right)}{\ln a}\right)}-\frac{\ln\left(\frac{1}{\ln a}\right)}{\ln a}$$

$$=e^{\ln\left(\frac{1}{\ln a}\right)}-\frac{\ln\left(\frac{1}{\ln a}\right)}{\ln a}$$

$$=\frac{1}{\ln a}-\frac{\ln\left(\frac{1}{\ln a}\right)}{\ln a}$$

Then using $g(x_0)=0$,

$$\ln\left(\frac{1}{\ln a}\right)=1$$ $\implies$ $$\ln a=e^{-1}$$ $\implies$ $$a=e^{e^{-1}}$$

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Now, substitute these values of $a$ and $x_0$ back into $g$ and $g'$ to indeed show $g(x_0)=g'(x_0)=0$ and hence the assumption is true.

Finally,

$$g''(x)=(\ln a)^2 a^x>0 \,\,\,\forall x\in\mathbb R$$

and hence $g$ is convex and $x_0$ is the global minimum. This shows (1) and (2).

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In fact, as we've shown $((\exists \,\,a>1,x_0>0)\,\,g(x_0)=g'(x_0)=0) \iff \left(a=e^{e^-1}\right) \land \left(x_0=\frac{\ln\left(\frac{1}{\ln a}\right)}{\ln a}=e\right)$, we can also conclude this $a$ is unique.

Answer 2

Let $f_a(x)=a^x-x$. Then $$f'_a(x)=a^x\ln a-1$$ and $$f''_a(x)=a^x(\ln a)^2>0$$ For $a>1$, the first derivative vanishes at $$x=-\log_a\ln a=-\frac{\ln\ln a}{\ln a}$$

Now define $$g(a)=a^{-\ln\ln a/\ln a}+\frac{\ln \ln a}{\ln a}$$

And apply Bolzano's theorem to prove that $g$ vanishes at some point $c$.

Then $f_c(-\ln\ln c/\ln c)=f'_c(-\ln\ln c/\ln c)=0$. So $f_c$ has a minimum at the point $(-\ln\ln c/\ln c,0)$.

Answer 3

Solution

Let $a>1$ and $$ a^x = x $$

hence,

$$ x^{1/x} = a $$

To see the graph of $x^{1/x}$, follow this Desmos link below,

https://www.desmos.com/calculator/nwtxfbjwgu

Notice that the graph contains only $1$ maxima.

Hence only the value of $a$ for which $x^{1/x}$ is maximum will have only one real root for the equation $a^x = x$.

All other values of $a$ will give either $0$ or $2$ solutions.

Finding that special value of $a$ is easy, just differentiate $x^{1/x}$ and set it to zero.

It comes out to be,

$a = e^{1/e}$

Appendix : How to calculate the derivative of $x^{1/x}$?

$$ y = x^{1/x} $$

Taking the natural logarithm on both sides,

$$ \ln{y} = \frac{1}{x}\ln{x} $$

Now taking the derivative of both sides and using the chain rule,

$$ \frac{1}{y}*\frac{dy}{dx} = \frac{-1}{x^2} + \frac{\ln{x}}{x^2} $$

Setting $\frac{dy}{dx} = o$ and rearranging we get,

$$ \ln{x} = 1 $$

It follows that,

$$ x=e $$

and that the special value of $a$ is

$$ e^{1/e} $$

More generally we can also say that for $a<0$, there are $0$ solutions. For $a=0$, it is not defined. For $e^{1/e}>a>0$, there are exactly $2$ solutions. For $a=e^{1/e}$, there is exactly $1$ solution. For $a>e^{1/e}$, there is no solution.

Answer 4

Let $$f(x)=a^x-x$$with $a>1$, So $$f(0)=1>0,$$$$f(1)=a-1>0.$$ $$f'(x)=a^x*\ln a-1,$$$$f''(x)=a^x*(\ln a)^2>0.$$This implies $f'(x)$ increases everywhere. To see whether it has a zero, let $$a^\alpha*\ln(a)-1=0$$ and we get $$\alpha=-(\ln a)^{-1}*\ln(\ln a)\in \mathbb R.$$So $f(x)$ decreases for $x<\alpha$ and increases for $x>\alpha$. Thus it is enough to check whether $$f(\alpha)>0$$

Noticing if $a=e^{e^{-1}}$, $$f(\alpha)={e^{\alpha *e^{-1}}}-\alpha=0.$$

Let $g(a)=a^{\alpha}-\alpha$ with $a>1$, $g'(a)=\alpha *a^{\alpha-1}$, and $g'(a)>0$ for $1<a<e$, $g'(a)<0$ for $a>e$. $g(e)=1>0$, $g(+\infty)=(+\infty)^{\alpha}-{\alpha}>0$ because $\alpha<0$. So we know that $g(a) $ increases for $1<a<e$ with $g(e^{e^{-1}})=0$, and decreases for $a>e$ with no zeros in the right side. It is obvious that $g(a)$ can not have consecutive zeros on $\mathbb R$. Otherwise, let $g(t)=0$ hold for all $t\in (p,q)$ with $e>q>p>1$, then due to monotonicity, that holds also for all $t\in (1,q)$. But that cannot be true since $1$ is a singularity of $g(t)$ and $g$ is continuous on $(1, +\infty)$.

So $a^{\alpha}-\alpha>0$ if $a>e^{e^{-1}}$ and $a^{\alpha}-\alpha<0$ if $a<e^{e^{-1}}$.

In conclusion, if $1<a<e^{e^{-1}}$ then $a^x=x$ has two solutions; if $a=e^{e^{-1}}$ then $a^x=x$ has one solution; if $a>e^{e^{-1}}$ then $a^x=x$ has no solution.

Hope this could help you!

Answer 5

I guess I will add my two cents $$a^x=x$$ $$a=x^{{1}/{x}}$$ now $$e^{x}\ge 1+x$$ $$e^{(x-e)/{e}}\ge x/e$$ $$e^{x/e}\ge x$$ $$e^{1/e} \ge x^{1/x}$$

Therefore maximum of $a=e^{1/e}$ for solution. Thus there exists solution for $a\in\mathbb{(1,e^{1/e})}$