Question

How do you solve `log_6x + log_6(x+1) =1`?

Answer 1

`x=2`

Explanation:

The left side has two logs with same base x, hence these two can be combined to give `log_6 x(x+1) =1`

This implies `x(x+1)=6`,

`x^2 +x-6`=0
`(x+3)(x-2)=0`
`x=-3" "`, `x=2`

But `x=-3` is an extraneous solution, because `log_6(-3)` is not a real number.