Significance of $1.32109977$ in the area of exponential/logarithmic functions I noticed that if you graph $y = log_n(x)$ and $y = x^n$ simultaneously for various values of $n$. Specifically, I noticed that with higher values of $n$, the graphs crossed twice. For lower values of $n$ (close to $1$) the graphs did not cross. I wondered at what point there would be exactly one shared point. I figured this out by using Wolfram Alpha. I inputted log base $n(x) = x^n$ where $n=<number>$, and tried to make the outputted solutions as close to each other as I can. I ended up with the number $1.32109977$. I'm curious as to 3 things:
1) What is the exact value of this number (if that can be found)?
2) Is this number at all relevant or just a value where a graph happens to look a certain way?
3) What is this number called? (...because out of a world of >7,000,000,000 people, someone has named this number)
 A: According to alpha, a solution to $$\log_n(x)=x^n$$ is $$\sqrt[n]{-\frac{W(-n\log(n))}{n\log(n)}}$$
Here $W$ denotes the Lambert W function. You can also read about it on alpha. The argument of the $n$'th root is actually positive, provided $-n\log(n)\ge -\frac{1}{e}$ (apart from a removable singularity at $n=1$).  If the input to $W$ is less than $-1/e$, the result is non-real, and hence the solution will become complex.  Hence the critical value you seek is $$n\log n=1/e$$
This has exact value $$e^{W(1/e)}\approx 1.321099762015617456962355870878829561624$$
As far as I know this constant has no particular significance.
A: If you want the equation $$\log_n(x)=x^n$$ to have precisely one solution in the manner you describe, then since the left quantity is concave down, and the right is concave up, then you want the two quantities to have equal values at the same place where they have equal derivatives. So you want to solve the system $$\begin{cases}\frac{\ln(x)}{\ln(n)}=x^n\\\frac{1}{x\ln(n)}=nx^{n-1}\end{cases}$$
The second equation can be used to solve for $x$. 
$$x=\left(\frac{1}{n\ln(n)}\right)^{1/n}$$
Then you can use this in the first equation to give 
$$\begin{align}
\frac{\ln\left(\frac{1}{n\ln(n)}\right)}{n\ln(n)}-\left(\frac{1}{n\ln(n)}\right)
&=0\\
\ln\left(\frac{1}{n\ln(n)}\right)-1
&=0\\
\ln\left(\frac{1}{n\ln(n)}\right)
&=1\\
n\ln(n)
&=e^{-1}\\
W(n\ln(n))
&=W(e^{-1})\\
\ln(n)
&=W(e^{-1})\\
n&=e^{W(e^{-1})}\\
n&\approx1.321099762015617456962355870878829561623557500156399864816\ldots
\end{align}$$
Where $W$ is the Lambert $W$ function. It is an inverse function for $z\mapsto ze^z$. This means $W(ze^z)=z$, and with $z=\ln(n)$, you have $W(n\ln(n))=\ln(n)$. And this relation is used in the above equations.
