Solve for $x = 8\ln(x)$, with $x > 0$

There are 2 intersections between $f(x) = x$ and $g(x) = 8\ln(x)$. 1

I tried solve it by replacing the logarithm with an exponential:

$$\begin{align*} 8\ln(x) = x \\ e^{8\ln(x)} = e^x \\ x^8 = e^x \\ x^{\frac{8}{x}} = e \\ \end{align*}$$

But now I'm stuck. What do I need to learn in order to solve it? Or what did I miss?

Thank you for your help!

Solution 1:

Solution of this is not possible in "elementary functions". However, it can be done using the Lambert W function.

Definition: $x=ye^y$ iff $W(x) = y$.
So, in this case $$ x = 8\ln x \\ e^x = x^8 \\ e^{x/8} = x \\ 1=xe^{-x/8} \\ -\frac{1}{8} = \left(-\frac{x}{8}\right)e^{-x/8} \\ W\left(-\frac{1}{8}\right) = -\frac{x}{8} \\ -8 W\left(-\frac{1}{8}\right) =x $$

Using the two real branches of the $W$ function, our solutions are $$ x=1.155370825\quad \text{ and }\quad x=26.09348548 $$

Solution 2:

If you are searching for an explicit solution you cannot solve this properly as, as you 've seen, taking the exponential to suppress the log will create an exponential that you need to take the log to suppress ... becoming an infinite loop

If you are searching an approximation you can use Newton-Raphson's procedure however if you are searching for a way to express it, you can check the Lambert Function