Finding values for $a >0$ where $y=x$ intersects $y=a^x$ So the full question goes as following:
For what values $a>0$ does the curve $y=a^x$ intersect the straight line $y=x$?
I'm not really sure how to approach this differently than finding the values for x which satisfies $x=a^x$, thereafter would I find a. It was not as straight forward as I would imagine.
Any suggestions?
 A: The case $a \leq 1$ is easy, so let us suppose $a > 1$. Now look at the function $f: (0, \infty) \rightarrow \mathbb{R}$
$$
f(x) = \frac{a^x}{x}.
$$
Our goal is to find all the minima of $f$. We have
$$
f'(x) = \frac{a^x(x \ln(a) - 1)}{x^2},
$$
so $f$ has a local minimum at $x = \frac{1}{\ln(a)}$. One can check that this is in fact a global minimum. 
Now note that $a^x$ and $x$ intersect for some $x > 0$ if and only if $f(x) \leq 1$ for some $x > 0$ if and only if $f\left(\frac{1}{\ln(a)}\right) \leq 1$. But we have
$$
f\left(\frac{1}{\ln(a)}\right) = \ln(a) \cdot a^{\frac{1}{\ln(a)}} = \ln(a) \cdot e,
$$
so we want $\ln(a) \leq \frac{1}{e}$, i.e. $a \leq e^{\frac{1}{e}}$.
A: consider the function $$f(x)=x-a^x$$ then $$f'(x)=1-a^x\ln(a)$$ conyou finish now?
solve the equation $$1=a^x \ln(a)$$ for $x$, and note that $$f''(x)=-a^x(\ln(a))^2<0$$ for all real $x$
and note that for $$0<a<1$$ is $$f'(x)>0$$
A: The line $y = x$ intersects the curve $y = a^x$ ($a > 0$) when
$$x = a^x.$$
This equation can be solved for $x$ in terms of the Lambert W function as follows. Rearranging we have
\begin{align*}
x &= a^x\\
x &= e^{x \ln a}\\
x e^{-x \ln a} &= 1\\
-x \ln a e^{-x \ln a} &= -\ln a,
\end{align*}
and since this last equation is now exactly in the form for the defining equation for the Lambert W function, namely
$$\text{W} (x) e^{\text{W} (x)} = x,$$
where $\text{W} (x)$ denotes the Lambert W function, we have
$$-x \ln a = \text{W}_\nu (-\ln a),$$
or
$$x = -\frac{1}{\ln a} \text{W}_\nu (-\ln a).$$
Here $\nu$ denotes the two real branches for the Lambert W function ($\nu = 0, -1$). When $a > 1$, $\ln a$ will be positive. So for two real solutions both branches for the Lambert W function are selected with the argument for the Lambert W function lying between $-1/e$ and zero. Thus
$$-\frac{1}{e} \leqslant -\ln a < 0,$$
or
$$1 < a \leqslant e^{1/e}.$$
Note that graphically one has two points of intersection between the curve and the line (one point only when $a = e^{1/e}$).
When $0 < a < 1$, as $\ln a$ will be negative the argument for the Lambert W function is positive meaning the principal branch ($\nu = 0$) is selected. Thus there will be only one point of intersection between the curve and the line when $0 < a < 1$.
The case $a = 1$ is trivial. So the values of $a > 0$ for which the curve $y = a^x$ intersects the line $y = x$ are $0 < a \leqslant e^{1/e}$. 
A: If the two curves intersect, they do at $x>0$ where $x=a^x$, which is the same as $x\log a-\log x=0$.
The function $f(x)=x\log a-\log x$, defined for $x>0$, has derivative
$$
f'(x)=\log a-\frac{1}{x}
$$
Note that, when $0<a\le 1$, the derivative is everywhere negative. Since we have
$$
\lim_{x\to0}f(x)=\infty,\qquad f(1)=\log a<0 \qquad(0<a\le 1)
$$
the equation $f(x)=0$ has a single solution, which is in the interval $(0,1)$ (blue curve, $a=1/2$) except for $a=1$ where the solution is $x=1$.
Assuming $a>1$, the derivative vanishes at $x=1/\log a$, which is an absolute minimum. Since
$$
f(1/\log a)=1-\log(1/\log a)=1+\log\log a
$$
there will be


*

*no solution for $1+\log\log a>0$, that is, $a>e^{1/e}$ (black curve, $a=2$);

*one solution for $a=e^{1/e}$ (orange curve);

*two solutions for $1<a<e^{1/e}$ (red curve, $a=1.2$).



Note: the drawn curves are just examples of what happens in the various cases.
