Determine all positive numbers $a$ for which the curve $y = a^x$ intersects the line $y = x$ without calculus The answer is $0 < a < e^{1/e}$ , but how to find it? Is it a system of equations? Which ones? I just need an idea at least, because I'm stuck.
If it is impossible without calculus, solve it with it.
 A: We have
$$
a^x = x \implies x \ln(a) = \ln(x) \implies \frac{\ln(x)}{x} = \ln(a)
$$
This equation is indeed unsolvable without some amount of calculus.
Using calculus, however, we may show that $\ln(x)/x$ is a function that increases from $-\infty$ at $x = 0$, reaches its maximum of $\frac 1e$ at $x = e$, and then decreases as $x \to \infty$ towards $0$.
With that in mind, it is clear that the curve $y = \ln(a)$ will intersect the curve $y = \ln(x)/x$ if and only if $-\infty < \ln(a) \leq \frac 1e$, which is to say that $0 < a \leq e^{1/e}$.
A: This is just a complement to previous answers.
As already told by Omnomnomnom $$a^x = x \implies x \ln(a) = \ln(x) \implies \frac{\ln(x)}{x} = \ln(a)$$ connot be solved in terms of elementary functions. The only explicit solution is given using Lambert function $$x=-\frac{W(-\log (a))}{\log (a)}$$ which has been worked by Euler too; in particular, Euler showed that this equation has a vertical asymptote corresponding to $a=e^{\frac{1}{e}}$
A: Part of this can be done without calculus. Consider $a<1$, $a=1$, $a>1$ as separate cases. The first two can be done without calculus. $a=1$ is simple, the intersection is at $x=1$. For $a<1$, $f(x)=a^x - x$ becomes large and positive as you go to $-\infty$ (the exponential dominates) and large and negative as you go to $+\infty$ (the exponential decays and the linear term dominates). So it must change sign, so it must be zero somewhere. 
For $a>1$, $f(x)$ is large and positive as you go to $+\infty$ or to $-\infty$, so you can't do the method above. Instead, since we know the function is positive at "most" points, we want to find its minimum. We can't do this without calculus. Omnomnomnom has already shown how you would do it with calculus.
