Solve the equation.
-$\log _ { 4 } ( x + 2 ) = 2 + \log _ { 4 } ( x - 4 )$
A) $\left\{ - \frac { 22 } { 5 } \right\}$
B) $\left\{ - \frac { 2 } { 5 } \right\}$
C) $\left\{ \frac { 2 } { 5 } \right\}$
D) $\left\{ \frac { 22 } { 5 } \right\}$

Question

Quizplus · Accepted Answer

To solve the equation $\log _ { 4 } ( x + 2 ) = 2 + \log _ { 4 } ( x - 4 )$, we first isolate the logarithmic expressions: $\log _ { 4 } ( x + 2 ) - \log _ { 4 } ( x - 4 ) = 2$. Using the property of logarithms that $\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})$, we get $\log _ { 4 } \left(\frac{x + 2}{x - 4}\right) = 2$. To remove the logarithm, we rewrite the equation in exponential form: $4^2 = \frac{x + 2}{x - 4}$. Simplifying, we get $16 = \frac{x + 2}{x - 4}$, which leads to $16(x - 4) = x + 2$. Solving for $x$, we find $x = \frac{22}{5}$.

Solve the Equation log⁡4(x+2)=2+log⁡4(x−4)\log _ { 4 } ( x + 2 ) = 2 + \log _ { 4 } ( x - 4 )log4​(x+2)=2+log4​(x−4)

Solve the Equation $\log _ { 4 } ( x + 2 ) = 2 + \log _ { 4 } ( x - 4 )$