Question

How do you solve `log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1`?

Answer 1

`x = 1/6 (1 + sqrt[85])`

Explanation:

`log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1` or
`log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = log_(1/3)(1/3)^(-1)`
then

`(x^2 + 4x)/ (x^3 - x) =3` or `(x + 4)/ (x^2 - 1) =3`

or

`3x^2-3=x+4` or finally `3x^2-x-7=0`

solving for `x` we get `{x = 1/6 (1 - sqrt[85])}, {x = 1/6 (1 + sqrt[85])}` but we choose `x = 1/6 (1 + sqrt[85])` because `x = 1/6 (1 - sqrt[85]) <0` and makes `(x^3 - x) < 0`