How to solve: log_(3) (x+6) + log_(3)(x-6) = 4

Question

How to solve:
`log_(3) (x+6) + log_(3)(x-6) = 4`

Answer 1

Collapse the logarithms into one:

`\log_3(x+6)+\log_3(x-6)=\log_3((x+6)(x-6))=4`

Then write both sides as powers of 3:

`3^(\log_3((x+6)(x-6)))=3^4`

`(x+6)(x-6)=81`

Simplify the left side and solve for
`x`:

`x^2-36=81`

`x^2=117`

`x=\pm√(117)=\pm3√(13)`

But we're not done yet. Notice that both
`x+6` and
`x-6` are negative when
`x=-3√(13)`. We can't take the logarithm of a negative number (the result isn't real-valued, anyway), so we throw this solution out, and we're left with just

`x=3√(13)`