Originally Posted by

**Soroban** Hello, DannyMath!

What you did was correct . . .

Note that: .$\displaystyle x \ne 0$

We have: .$\displaystyle \log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x) $

. . . . . . . . . . $\displaystyle 6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)$

. . . . . . . . . . .$\displaystyle \log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x $

Then: .$\displaystyle x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0$

Therefore: . $\displaystyle \rlap{//////}x \,=\,0,\;\;x\,=\,90$