Question

How do you solve `log[(x + 3)(x - 8)] + log[(x + 3)/(x - 8)] = 2`?

Answer 1

`S = -13`

Explanation:

By the rules of logarithms, `log(m) + log(n) = log(mn)`, so

`log[(x+3)(x-8)] + log[(x+3)/(x-8)] = log[((x+3)(x-8)(x+3))/(x-8)]`

We cancel the (x-8) on the numerator and the denominator

`log[(x+3)(x-8)] + log[(x+3)/(x-8)] = log[(x+3)^2]`

The problem told us that that equaled 2, so

`log[(x+3)^2] = 2`

We know that if `log_b(a) = c` then `a = b^c` so we have

`(x+3)^2 = 10^2`

We use 10 because if the log doesn't have a base, it's implicitly 10. (Also note that we couldn't have put that square to the front of the logarithm and cut it with the 2 because we'd lose a root if we did it)

Now we take the square root,

`x + 3 = +-10`

And solve these values
`x = 10 - 3 = 7`
`x = -10 - 3 = -13`

But before that, check for the values we know x can't be! Logarithms can't take null or negative arguments so we know `x != 3 and x!=8`

We know that `x-3` and `x-8` can only be negative if the other is negative too. so we study the signs

------(-3)++++++++++++ `(x-3)`
----------------------(+8)++ `(x-8)`

++++0---------------0+++ `(x-3)(x-8) or (x-3)/(x-8)`

We know we can't take values that are 0 or negative, so `x > 8 or x < -3` which means there's only one possible root, -13.