Have We Met Before? No, Just This Once.

Since the equation has a unique solution, then at the point of contact of two graphs, their derivatives must be same i.e

$\displaystyle a^x\ln a=1 (*)$

Taking natural logarithm on both the sides of original equation: $\Rightarrow x\ln a=\ln x$ and differentiating wrt $x$ , $\ln a=\dfrac{1}{x} \Rightarrow x\ln a=1 \Rightarrow \ln a^x=1 \Rightarrow a^x=e$ . Combining this with $(*)$ ,

$\displaystyle e\ln a=1 \Rightarrow \boxed{a=e^{1/e}}$

I'll write up a full solution later when I have time, but for now, observe that if $a = e^{\frac{1}{e}}$ then $a^{e} = e$ . Note that this value of $a$ exceeds $1$ , so we can have the graphs $y = a^{x}$ and $y = x$ touch for some $a \gt 1$ .

This note was to address someone's request for clarification. It turns out that this value of $a$ is the one we're after; stay tuned for a detailed proof.

EDIT: O.k., now for my solution, as promised.

With $a \gt 1$ let $f(x) = a^{x} - x$ and suppose that $f(x) = 0$ has a unique solution.

Now $f'(x) = a^{x}*\ln{a} - 1$ and $f''(x) = a^{x}*(\ln{a})^{2}$ .

The critical point for $f(x)$ is where $f'(x) = 0 \Longrightarrow x = -\dfrac{\ln{(\ln{a})}}{\ln{a}}$ .

Next, note that $f''(x) \gt 0$ for all $x$ for any $a \gt 1$ we choose, and so the critical point we found above represents an absolute minimum. Since we supposed that $f(x) = 0$ has a unique solution, this unique solution must be at the critical point. So

$a^{-\frac{\ln{(\ln{a})}}{\ln{a}}} = -\dfrac{\ln{(\ln{a})}}{\ln{a}}$

$\Longrightarrow e^{-\ln{(\ln{a})}} = -\dfrac{\ln{(\ln{a})}}{\ln{a}}$ , (since $a = e^{\ln{a}}$ ),

$\Longrightarrow \dfrac{1}{\ln{a}} = -\dfrac{\ln{(\ln{a})}}{\ln(a)}$

$\Longrightarrow \ln{(\ln{a})} = -1 \Longrightarrow \ln{a} = \dfrac{1}{e} \Longrightarrow a = e^{\frac{1}{e}}$ .

So if $f(x) = 0$ is to have a unique solution then this will be the value of $a$ . With this value of $a$ we have the critical point $x = e$ , for which $a^{x} = x$ , allowing us to conclude that the solution is indeed $a = \boxed{e^{\frac{1}{e}}}$ .

EDIT #2: Since $1000*a = 1444.6679...$ , we have $\lceil 1000*a \rceil = \boxed{1445}$ as the answer required for entry. Sorry I didn't specify that in my original solution.

The " $\lceil x \rceil$ " symbols indicate the ceiling function, which returns the smallest integer $\ge x$ .

$a^x=x$

$x\ln a=\ln x$

$\frac{\ln x}x=\ln a$

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Let $y=\frac{\ln x}x$ . The question is equivalent to finding $y$ which corresponds to only one value of $x$ .

Let us differentiate $y$ with respect to $x$ .

$\frac{\mathrm dy}{\mathrm dx} = \frac{\left(\frac{\mathrm d}{\mathrm dx}\ln x\right)x - \ln x\left(\frac{\mathrm d}{\mathrm dx}x\right)}{x^2} = \frac{1-\ln x}{x^2}$

Its local extrema: $\frac{1-\ln x}{x^2}=0 \implies 1-\ln x=0 \implies \ln x=1 \implies x=e$ .

When $x=e$ , $y=\frac{\ln e}e=\frac1e$ .

It is easy to verify that this is the maximum.

Also, $\lim_{x\to\infty}y=0$ , which means that when $x>e$ , $0<y<\frac1e$ .

Also, $\lim_{x\to0^+}y=-\infty$ , which means that this is how the graph of $y$ to $x$ looks like:

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Beginning at when $x>0$ (the other values of $x$ lead to an undefined value of $y$ ), we start at infinity, rapidly increasing to $y=\frac1e$ when $x=e$ , and then gradually decreasing, approaching zero, but never reaching it.

Therefore $y\le0$ and $y=\frac1e$ would correspond to only one value of $x$ .

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For the first case:

$y\le0$

$\ln a\le0$

$a\le1$ (rejected)

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For the second case:

$y=\frac1e$

$\ln a=\frac1e$

$a=e^{\frac1e}$

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$\lceil1000a\rceil=\lceil1444.6\cdots\rceil=1445$ .

"This problem, once solved, becomes simple" is something my father always had up in his office, in a picture frame. Sometimes, if you have a hunch what the answer might be, the proof isn't very difficult. Let's say that the hunch is that $a={ e }^{ \frac { 1 }{ e } }$ . First, we differentiate to find the point of tangency

$\dfrac { d }{ dx } (a^{ \displaystyle x }-x)={ a }^{ \displaystyle x }Log(a)-1=0$

from which both of these follow

$\displaystyle{ a }^{ \displaystyle x }=\dfrac { 1 }{ Log(a) }$

$\displaystyle{ a }^{ \displaystyle{ a }^{\displaystyle x } }=e$

But we want this point of tangency at $0$ , that is

$\displaystyle{ a }^{ \displaystyle x }=x$ , which means

$\displaystyle{ a }^{ \displaystyle{ a }^{ \dfrac { 1 }{ Log[a] } } }=e$

If one plugs in $\displaystyle a={ e }^{ \frac { 1 }{ e } }$ (because it's such a great guess), then that's exactly what we get, which is $e$ .

Consider the function $f_a(x) = a^x - x$ . We then have $f_a^\prime(x) = a^x\log a - 1$ and $f_a^{\prime\prime} = a^x \log^2 a$ . For $a>1$ , its graph is convex (second derivative ${}>0$ ). We want to find $a$ such that $f_a$ has one root. If $f_a$ has one root, then that is also where its minimum is. Find its minimum by solving $f_a^\prime(x) = 0$ : $a^x\log a - 1 = 0$ ; hence, $x = \log_a(1/\log{a})$ . Now, find $a$ such that the minimum equals 0; that is, solve $f_a(\log_a(1/\log{a})) = 0$ . We have $f_a(\log_a(1/\log{a})) = a^{\log_a(1/\log{a})} - \log_a(1/\log{a}) =$ $1/\log{a} - \log(1/\log{a})/\log a = (1 - \log(1/\log{a})) / \log{a}$ Thus, the minimum equals 0 for $1 = \log(1/\log{a})$ . Hence, for $a = e^{1/e}$ . (No need for guessing :-)

The question should be:

Let $a$ be the largest real number such that $a^x=x$ is true. Compute $\left\lceil 1000a \right\rceil$ .

As most people did, I viewed the line $y=x$ as a tangent and hence the derivative of ${ a }^{ x }=x$ can be set to 1. ${ a }^{ x }\ln { a }=1$ . ${ a }^{ x }$ is then replaced with $x$ . $x\ln{ a }=1\Rightarrow$ $a={ e }^{ \frac { 1 }{ x } }$ Substituting this value of a into the original expression solves for $x=e\Rightarrow$ $a={e}^{\frac{1}{e}}$