How many solutions: $ a^{x} = x $

Question

I'm clueless.

For $ a > 0 $ how many solutions: $ a^{x} = x $

Answer 1

Since $a^x>0$, for all $x$, if a solution exists it is positive. So it's not restrictive to assume $x>0$. Then the equation becomes the same as $$ x\log a=\log x $$ (natural logarithm). Let's look at $a>1$, to begin with.

Consider the function $f(x)=x\log a-\log x$. It's easy to see that $$ \lim_{x\to0}f(x)=\infty, \qquad \lim_{x\to\infty}f(x)=\infty $$ Thus the function has an absolute minimum. Since $$ f'(x)=\log a-\frac{1}{x} $$ the minimum is at $\frac{1}{\log a}$. Now $$ f\left(\frac{1}{\log a}\right)=1+\log\log a $$ and we have

• no solution if $1+\log\log a>0$, that is, $a>e^{1/e}\approx 1.44467$;
• one solution if $1+\log\log a=0$, that is, $a=e^{1/e}$;
• two solutions if $1+\log\log a<0$, that is, $1<a<e^{1/e}$.

The case $a=1$ is obvious.

In the case $0<a<1$, the function $f$ has $$ \lim_{x\to0}f(x)=\infty, \qquad \lim_{x\to\infty}f(x)=-\infty $$ and $f'(x)<0$ for all $x>0$, so the equation has only one solution.

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Alternative method. The equation, with the same trick, is equivalent to $$ \frac{\log x}{x}=\log a $$ Let's consider the function $g(x)=\frac{\log x}{x}$. Then $$ \lim_{x\to 0}g(x)=-\infty, \qquad \lim_{x\to\infty}g(x)=0 $$ Since $$ g'(x)=\frac{1-\log x}{x^2} $$ we see that $g$ has a maximum at $x=e$; moreover $g(e)=\frac{1}{e}$.

Since $\log a$ takes on every real value exactly once, we see we have

• no solution if $\log a>1/e$
• one solution if $\log a=1/e$
• two solutions if $0<\log a<1/e$
• one solution if $\log a\le 0$

Answer 2

Assume $a>0$

$$a^x=x\Longleftrightarrow\ln\left(a^x\right)=\ln(x)\Longleftrightarrow$$ $$x\ln\left(a\right)=\ln(x)\Longleftrightarrow\ln\left(a\right)=\frac{\ln(x)}{x}\Longleftrightarrow$$ $$a=e^{\frac{\ln(x)}{x}}\Longleftrightarrow a=\exp\left[\frac{\ln(x)}{x}\right]\Longleftrightarrow a=\sqrt[x]{x}$$