Calculating $a$ for which the graph of $y=a^x$ touches the graph of $y=x$

Question

In the xy plane, the graph of $y = a^x$ touches the graph of $y=x$ iff there is a value $x$ for which $y$ has the same value in both cases ($a^x = x$) and they have equal first derivatives ($\mathrm{ln}(a) \cdot a^x = 1$). So my goal is to get the value of $a$ from this set of equations:

$$ a^x = x $$ $$ \mathrm{ln}(a) \cdot a^x = 1 $$

Assume that $a$ is positive. I experimentally found that $a$ is greater than $1.4$ and smaller than $1.5$.

I discovered that I need the Lambert W function to solve it. With a rule from the linked Wikipedia article, I found that the first equation can be transformed thus: $$ x = \mathrm{e}^{\mathrm{ln}(a) \cdot x} $$ $$ x = \frac{1}{\mathrm{ln}(a)} \mathrm{W} \left( -\mathrm{ln}(a) \cdot \mathrm{e}^{\mathrm{ln}(a)} \right) $$ $$ x = \frac{\mathrm{W} \left( -\mathrm{ln}(a) \cdot a \right)}{\mathrm{ln}(a)} $$

I didn't know what to do with the second equation, so I put it into Wolfram Alpha and believed that the solution is correct. The solution was this: $$ x = \frac{\mathrm{ln} \left( \frac{1}{\mathrm{ln}(a)} \right)}{\mathrm{ln}(a)} $$

The left sides of both new equations are equal, so their right sides are equal too, so I have this: $$ \frac{\mathrm{W} \left( -\mathrm{ln}(a) \cdot a \right)}{\mathrm{ln}(a)} = \frac{\mathrm{ln} \left( \frac{1}{\mathrm{ln}(a)} \right)}{\mathrm{ln}(a)} $$

I tried to solve it, but I don't know what to do with the W function and so many logarithms. How can it be solved?

Answer 1

This problem can be solved without using the Lambert W function,

suppose,the curves $y=a^x$,$y=x$,touch at the point $(b,b)$

so we have,$b=a^b$...(1) [since this point lies on $y=a^x$]

from the equation $a^x\ln a=1$,at $x=b$,we have,

$a^b\ln a =1$

from eq(1),$a^b=b$ so,

$b\ln a=1$

$\implies b=\frac{1}{\ln a}$

substituing in eq(1),

$ \frac{1}{\ln a}=a^\frac{1}{\ln a}$

$\frac{1}{\ln a}=e$

$\ln a =\frac{1}{e}$

$\implies a=e^{1/e}$

Answer 2

Your $W$-expression is slightly wrong; it should involve $W(-\ln a)$. But after that, the key is to note that around $1.4\le a\le1.5$, $W(-\ln a)$ is multivalued because $-\ln a\in(-1/e,0)$. There is only a single value where the two corresponding branches meet, i.e. $-a\ln a=-\frac1e$. Solving this gives $a=e^{1/e}$.