Question

How do you solve `log(x+3)+log(x-3)=log27`?

Answer 1

`x = 6`

Explanation:

First of all, this equation is defined on `]3,+oo[` because you need `x+3 > 0` and `x - 3 > 0` at the same time or the log won't be defined.

The log function maps a sum into a product, hence `log(x+3) + log(x-3) = 27 iff log [(x+3)(x-3)] = log 27`.

You now apply the exponential function on both sides of the equation : `log [(x+3)(x-3)] = log 27 iff (x+3)(x-3) = 27 iff x^2 - 9 = 27 iff x^2 - 36 = 30`. This is a quadratic equation that has 2 real roots because `Delta = -4*(-36) = 144 > 0`

You know apply the quadratic formula `x = (-b+- sqrtDelta)/2a` with `a = 1` and `b = 0`, hence the 2 solutions of this equation : `x = ± 6`

`-6 !in ]3,+oo[` so we can't keep this one. The only solution is `x = 6`.