im not sure how to type logs so try to bare with me >< first set of parenthesis is the base of the log
log(4)x + log(4)(x-2)= log(4)8
if you could explain that i would be super happy
Hello, Porcelain!
On the left side, combine the logs . . .$\displaystyle \log_4(x) + \log_4(x-2) \:=\:\log_4(8)$
We have: .$\displaystyle \log_4\bigg[x(x-2)\bigg] \:=\:\log_4(8)$
"Un-log" both sides: .$\displaystyle x(x-2) \:=\:8 \quad\Rightarrow\quad x^2 - 2x - 8 \:=\:0$
. . Then: .$\displaystyle (x-4)(x+2)\:=\:0 \quad\Rightarrow\quad x \:=\:4,-2$
But $\displaystyle x = -2$ is an extraneous root.
. . Therefore, the solution is: .$\displaystyle \boxed{x \:=\:4}$