# What is x if log_4 x=2 - log_4 (x+6)?

Jun 8, 2018

See process below

#### Explanation:

In this type of equations, our goal is to arrive to an expresion like
${\log}_{b} A = {\log}_{b} C$ from this, we conclude $A = C$

Lets see

${\log}_{4} x = 2 - {\log}_{4} \left(x + 6\right)$

We know that $2 = {\log}_{4} 16$. then

${\log}_{4} x = {\log}_{4} 16 - {\log}_{4} \left(x + 6\right) = {\log}_{4} \left(\frac{16}{x + 6}\right)$ applying the rule

$\log \left(\frac{A}{B}\right) = \log A - \log B$

So, we have $x = \frac{16}{x + 6}$

${x}^{2} + 6 x - 16 = 0$

$x = \frac{- 6 \pm \sqrt{36 + 64}}{2} = \frac{- 6 \pm 10}{2}$

Solutions are ${x}_{1} = - 8$ and ${x}_{2} = 2$ we reject negative and the only valid solution is ${x}_{2}$

Jun 8, 2018

$x = 2$

#### Explanation:

$\text{using the "color(blue)"laws of logarithms}$

•color(white)(x)logx+logy=log(xy)

•color(white)(x)log_b x=nhArrx=b^n

$\text{ add "log_4(x+6)" to both sides}$

${\log}_{4} x + {\log}_{4} \left(x + 6\right) = 2$

${\log}_{4} x \left(x + 6\right) = 2$

$x \left(x + 6\right) = {4}^{2} = 16$

${x}^{2} + 6 x - 16 = 0$

$\left(x + 8\right) \left(x - 2\right) = 0$

$x = - 8 \text{ or } x = 2$

$x > 0 \text{ and } x + 6 > 0$

$\text{thus "x=-8" is invalid}$

$\Rightarrow x = 2 \text{ is the solution}$