# How do you solve Log_2 X + Log_2 (X-3) = Log_2 (X^2-12)?

Dec 16, 2015

Use the $\log$ multiplication rule to determine
$\textcolor{w h i t e}{\text{XXX}} x = 4$

#### Explanation:

Log Multiplication Rule:
$\textcolor{w h i t e}{\text{XXX}} {\log}_{b} \left(a\right) + {\log}_{b} \left(c\right) = {\log}_{b} \left(a \cdot c\right)$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {\log}_{2} \left(x\right) + {\log}_{2} \left(x - 3\right) = {\log}_{2} \left(x \cdot \left(x - 3\right)\right) = {\log}_{2} \left({x}^{2} - 3 x\right)$

and the given equation:
$\textcolor{w h i t e}{\text{XXX}} {\log}_{2} \left(x\right) + {\log}_{2} \left(x - 3\right) = {\log}_{2} \left({x}^{2} - 12\right)$
is equivalent to
$\textcolor{w h i t e}{\text{XXX}} {\log}_{2} \left({x}^{2} - 3 x\right) = {\log}_{2} \left({x}^{2} - 12\right)$

This implies
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 3 x = {x}^{2} - 12$

$\textcolor{w h i t e}{\text{XXX}} - 3 x = - 12$

$\textcolor{w h i t e}{\text{XXX}} x = 4$

Dec 16, 2015

$x = 4$

#### Explanation:

First of all, use the fact that for any base, the sum of two logarithms is the logarithm of the product of the arguments. In formulas:

$\log \left(x\right) + \log \left(y\right) = \log \left(x y\right)$

This means that, in your case, we have that

${\log}_{2} \left(x\right) + {\log}_{2} \left(x - 3\right) = {\log}_{2} \left(x \left(x - 3\right)\right) = {\log}_{2} \left({x}^{2} - 3 x\right)$

So, the equation becomes

${\log}_{2} \left({x}^{2} - 3 x\right) = {\log}_{2} \left({x}^{2} - 12\right)$

Now, the logarithm is an injective function. This means that two different numbers can't have the same logarithm. Again, in formulas, this means that

$\log \left(a\right) = \log \left(b\right) \setminus \iff a = b$

${\log}_{2} \left({x}^{2} - 3 x\right) = {\log}_{2} \left({x}^{2} - 12\right) \setminus \iff {x}^{2} - 3 x = {x}^{2} - 12$
$\cancel{{x}^{2}} - 3 x = \cancel{{x}^{2}} - 12 \setminus \implies 3 x = 12 \setminus \implies x = 4$