# where do exponential and logarithmic functions intersect?

If $0<a<1$, then the graphs of $y=a^x$ and $y=\log_a(x)$ intersect at some point $(t(a),t(a))$. Does this function $t(a)$ have any nice expression? How much do we know about this function, except that it is obviously increasing and stays between $0$ and $1$? It is a very natural question, so I guess people thought about it before.

EDIT: why do I think it is a natural question? When we teach students calculus, one way to introduce the $\log_a(x)$ function is as the inverse of $a^x=e^{x \ln a}$. And drawing the graphs of $a^x$ and $\log_a(x)$ for $0<a<1$ in one coordinate system pulls attention to their intersection point right away.

EDIT 2: Thanks to lhf and Wolfram Alpha, the graph of $a^x=\log_a(x)$ for $x,a$ between $0$ and $1$ can be seen here.

It shows that for $a<t(e^{-1})=e^{-e}\approx 0.065988$, we have THREE points of intersection of graphs $a^x$ and $\log_ax$ (see example of $a=0.03$ here). At $a=e^{-e}$, the two graphs are tangent to each other (see here).

However, one of the points of intersection can be seen as the "principal branch", which I denoted $t(a)$ above. It looks like it has an inflection point at $a=e^{-1}\approx 0.367879$.

EDIT 3: Interestingly, the graphs of $a^x$ and $\log_a(x)$ do intersect for some values of $a>1$, namely, if $1<a<e^{1/e}\approx 1.44467$: example for $a=1.3$.

• @Vishal Did you read the inequality in the question?
– anon
Commented Oct 5, 2013 at 2:18
• In other words, you want to plot the fixed points of $\log_a \log_a x$.
– lhf
Commented Oct 5, 2013 at 2:20
• "It is a very natural question." Fixed points and iterated exponentials in my experience do not occur together naturally. (Just saying.)
– anon
Commented Oct 5, 2013 at 2:20
• See a plot of $x-\log_y \log_y x =0$ here: wolframalpha.com/input/…
– lhf
Commented Oct 5, 2013 at 2:24
• @anon: it is the other way around: for every $x$ except $x=e^{-1}$ there are two values of $a$ such that $a^x=\log_a x$. Commented Oct 5, 2013 at 6:45

Okay, this turned out to be an easy one. Since $a^x$ and $\log_a(x)$ are inverse to each other, their point of intersection (if unique) lies on the line $y=x$, so in particular, that value of $x=t(a)$ will satisfy $x=a^x$. Therefore, $a=x^{1/x}$, and $t(a)$ is the inverse for this function on $[0,1]$. The function $y=x^{1/x}$ (see graph here) has maximum value $e^{1/e}$ at $x=e$, it approaches $1$ from above as $x\to\infty$. Thus the functions $a^x$ and $\log_a(x)$ have two intersection points when $1<a<e^{1/e}$, they touch each other at $x=a$ if $a=e^{1/e}$, and have no intersection points if $a>e^{1/e}$.
A "closed form formula" for $t(a)$ may be obtained through Lambert $W$ function: Since $x=a^x$, one has: $$1=x/a^x=x\cdot e^{-x \ln a},$$ and $$-\ln a= -x \ln a \cdot e^{-x \ln a},$$which means that $$W(-\ln a)=-x\ln a,$$ so that $$x=\dfrac{W(-\ln a)}{-\ln a}.$$
A graph of this function shows that it does indeed look like the inverse to $a=x^{1/x}$ on $[0,1]$ (see the graph in the EDIT 2 part).
Another question remains: what is the formula for the two "non-principal" intersection points when $0<a<e^{-e}$?
• (Full disclosure, I haven't absorbed your answer, but the following is almost always the right response to "Why did my use of the $W$ function not get all the solutions?".) You have use the $0^{\text{th}}$ branch of the $W$ function. The $-1$ branch is also real valued. plot of both Commented Sep 18, 2020 at 23:15