$\displaystyle x^{log(x)}\;=\;100 \cdot x$
You started out nicely
$\displaystyle log(x) \cdot log(x)\;=\;log(100 \cdot x)$
Then you rather wandered off.
$\displaystyle [log(x)]^{2}\;=\;log(x) + log(100)$
This is tantalizingly familiar.
$\displaystyle [log(x)]^{2}\;-\;log(x)\;-\;2\;=\;0$
It's Algebra I from here. Don't forget the ORIGINAL Domain.
It appeared to be a more interesting problem if these were logs base 10, so I assumed that they were and just threw that step in without discussion to see if you were paying attention. You were! Good work.
log(100x) = log(100) + log(x)
After that, why would I say just "100"? We're stuck with the logarithm function until it is treated properly.