Question

How do you solve `Log(x+2)+log(x-1)=1`?

Answer 1

Take exponent of both sides to get: `(x+2)(x-1) = 10`

Solve the quadratic and discard the spurious solution to get `x=3`

Explanation:

Note that we require `x+2 > 0` and `x-1 > 0` in order that `log(x+2)` and `log(x-1)` be defined.

This boils down to requiring `x > 1`.

Take exponent of both sides to find:

`10 = 10^1 = 10^(log(x+2)+log(x-1)) = 10^log(x+2)*10^log(x-1)`

`= (x+2)(x-1) = x^2+x-2`

Subtract `10` from both ends to get:

`0 = x^2+x-12 = (x+4)(x-3)`

Which gives us `x=-4` or `x=3`

Discard the spurious solution `x=-4` since `log(x+2)` and `log(x-1)` are undefined for `x=-4`.

So `x=3`