Question

How do you solve `log_2(x^2 - 10)= log_2 (3x)`?

Answer 1

`x=5`

Explanation:

`log_b p = log_b q`
`rArr p=q`

Therefore
`color(white)("XXX")log_2(x^2-10)=log_2(3x)`
`color(white)("XXX")rArr x^2-10=3x`

`color(white)("XXX")rarr x^2-3x-10=0`

`color(white)("XXX")rarr (x-5)(x+2)=0`

`color(white)("XXX")rarr x=5color(white)("XX")orcolor(white)("XX")x=-2`

But neither `log` is defined when `x=-2` and therefore this result is extraneous.