# How do you solve log_(5) (x-1) + log_(5) (x-2) - log_(5) (x+6) = 0?

Dec 26, 2015

Use logarithm rules for adding and subtracting logarithms and simplify first before solving. The step by step working is given below.

#### Explanation:

${\log}_{5} \left(x - 1\right) + {\log}_{5} \left(x - 2\right) - {\log}_{5} \left(x + 6\right) = 0$

Note: $\log \left(1\right) = 0$ for any base.

${\log}_{5} \left(x - 1\right) + {\log}_{5} \left(x - 2\right) - {\log}_{5} \left(x + 6\right) = {\log}_{5} \left(1\right)$
${\log}_{5} \left(\frac{\left(x - 1\right) \left(x - 2\right)}{x + 6}\right) = {\log}_{5} \left(1\right)$

Note: ${\log}_{b} \left(P\right) + {\log}_{b} \left(Q\right) = {\log}_{b} \left(P Q\right)$ and ${\log}_{b} \left(P\right) - {\log}_{b} \left(Q\right) = {\log}_{b} \left(\frac{P}{Q}\right)$

$\frac{\left(x - 1\right) \left(x - 2\right)}{x + 6} = 1$

Now to solve the equation.

Start by cross multiplying.

$\left(x - 1\right) \left(x - 2\right) = \left(x + 6\right)$ As you can see this removes the denominator and makes it easier for us to solve.

Now simplify
$\left(x - 1\right) \left(x - 2\right) = x \left(x - 2\right) - 1 \left(x - 2\right)$
$\left(x - 1\right) \left(x - 2\right) = {x}^{2} - 2 x - x + 2$
$\left(x - 1\right) \left(x - 2\right) = {x}^{2} - 3 x + 2$

Our problem now becomes

${x}^{2} - 3 x + 2 = x + 6$
Subtracting $x + 6$ from both the sides we get.
${x}^{2} - 3 x + 2 - x - 6 = 0$
${x}^{2} - 4 x - 4 = 0$

Solving for $x$ using the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Here $a = 1$, $b = - 4$ and $c = - 4$
$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 4\right)}}{2 \left(1\right)}$
$x = \frac{4 \pm \sqrt{16 + 16}}{2}$
$x = \frac{4 \pm \sqrt{16 \cdot 2}}{2}$
$x = \frac{4 \pm 4 \sqrt{2}}{2}$
$x = \left(2 \pm 2 s q t \left(2\right)\right)$
$x = 2 - 2 \sqrt{2}$ will give a negative number and won't satisfy the given equation. Therefore the solution is $x = 2 + 2 \sqrt{2}$