Solve the equation.
-$\log _{4}(x+4)+\log _{4}(x-2)=2$
A) $x=4$
B) $x=4, x=-6$
C) $x=5$
D) $x=-6$

Question

Quizplus · Accepted Answer

Using the property of logarithms that allows combining two logs with the same base into a single log by multiplying their arguments, we get $\log_{4}((x+4)(x-2)) = 2$. Simplifying inside the log gives $\log_{4}(x^2+2x-8) = 2$. Converting to exponential form, $4^2 = x^2 + 2x - 8$, which simplifies to $x^2 + 2x - 24 = 0$. Factoring, we find $(x+6)(x-4) = 0$, so $x = -6$ or $x = 4$. However, substituting $x = -6$ back into the original equation gives a log of a negative number, which is undefined. Therefore, the only solution is $x = 4$.

Solve the Equation log⁡4(x+4)+log⁡4(x−2)=2\log _{4}(x+4)+\log _{4}(x-2)=2log4​(x+4)+log4​(x−2)=2 A) x=4x=4x=4 B) x=4,x=−6x=4, x=-6x=4,x=−6 C) x=5x=5x=5 D) x=−6x=-6x=−6

Solve the Equation $\log _{4}(x+4)+\log _{4}(x-2)=2$ A) $x=4$ B) $x=4, x=-6$ C) $x=5$ D) $x=-6$