TITLE: How to solve the following equation involving an exponential function
QUESTION [7 upvotes]: How do you solve $10^x = x$? I'm not sure how to solve this algebraically. Using log functions wasn't enough.

REPLY [2 votes]: I can show you some interesting ways to find the answer.
Start with $x=\log_{10}(x)$.
Substitute this into itself to get $x=\log_{10}(\log_{10}(x))$
Repeat this infinitely:  $$x=\log_{10}(\log_{10}(\log_{10}(\dots\log_{10}(x)\dots)))$$
Try plugging in a random number for the $x$ inside the logarithms and use a calculator that can calculate complex numbers with logarithms to find the result.
Plugging in different numbers may produce different answers, but all answers should work to solve $x=10^x$.
For numerical reassurance (I used google)
$$\log(2)=0.30102999566$$
$$\log(\log(2))=-0.52139022765$$
$$\log(\log(\log(2)))=-0.2828+1.3643i$$
$$\log(\log(\log(\dots\log(2)\dots)))=-0.119193073+0.750583294i$$
When I try to plug this back into $x=10^x$, I get that it works, with a very small amount of error.
Also note that:
$$10^x=10^{x\pm\log_{10}(e)2\pi in},n=0,1,2,3,\dots$$
And using that, we get $$x=10^{x+\log_{10}(e)2\pi in}\implies x=\log_{10}(x)\mp\log_{10}(e)2\pi in$$
And putting this into our substitution method:
$$x=\log_{10}(\log_{10}(\dots\log_{10}(x)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in$$