# How do you solve Log (x +3)(x-2) = Log (7x-11)?

Nov 17, 2015

This equation has one solution $x = 5$

#### Explanation:

First we have to find the domain of this equation. Since $\log$ functions are only defined for positive real numbers we have to solve inequalities:

$\left(x + 3\right) \left(x - 2\right) > 0$ and $7 x - 11 > 0$

From first inequality we get: $x < - 3 \vee x > 2$

From the second we get: $x > 1 \frac{4}{7}$

So we can write, that the domain is: D=(2;+oo)

Now we can solve the equation:

$\log \left(x + 3\right) \left(x - 2\right) = \log \left(7 x - 11\right)$

Since the base of logarythm is the same on both sides (none is written, so I suppose it is $10$), we can skip $\log$ signs and solve equation:

$\left(x + 3\right) \left(x - 2\right) = 7 x - 11$

${x}^{2} + x - 6 = 7 x - 11$

${x}^{2} - 6 x + 5 = 0$

$\Delta = {6}^{2} - 4 \cdot 1 \cdot 5$

$\Delta = 36 - 20 = 16$

$\sqrt{\Delta} = 4$

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{6 - 4}{2} = 1$

${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a}$

${x}_{2} = \frac{6 + 4}{2}$

${x}_{2} = 5$

Since ${x}_{1} \notin D$, the only solution of this equation is ${x}_{2} = 5$