Question

Solve the equation below for x: logx+log(x-2)=log(3x)

Answer 1

Answer:

x=5

Step-by-step explanation:

Given equation is

$\log\left(x\right)+\log(x-2)=\log(3x)$

Apply formula: $\log A+\log B=\log\left(AB\right)$

$\log\left(x(x-2)\right)=\log(3x)$

$\log\left(x^2-2x)\right)=\log(3x)$

Since both sides have equal bases so we can drop the log

$x^2-2x=3x$

$x^2-2x-3x=0$

$x^2-5x=0$

$x(x-5)=0$

Which gives x=0 and x-5=0

or x=0 and x=5

Original problem $\log\left(x\right)+\log(x-2)=\log(3x)$ is not defined at x=0 henc only x=5 is the final answer.