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

Dec 17, 2015

$x = 6$

#### 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 $\log \left(x + 3\right) + \log \left(x - 3\right) = 27 \iff \log \left[\left(x + 3\right) \left(x - 3\right)\right] = \log 27$.

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

You know apply the quadratic formula $x = \frac{- b \pm \sqrt{\Delta}}{2} a$ 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 $x = 6$.