The solution set of the equation $\log _ { 4 } ( x - 3 ) + \log _ { 4 } ( x - 1 ) = \log _ { 4 } 8$ is

A) $\left\{ - \frac { 1 } { 3 } \right\}$
B){5}
C) $\left\{ - \frac { 1 } { 3 } , 1 \right\}$
D){3,1}
E){-3,-1}

Question

The solution set of the equation $\log _ { 4 } ( x - 3 ) + \log _ { 4 } ( x - 1 ) = \log _ { 4 } 8$ is

A) $\left\{ - \frac { 1 } { 3 } \right\}$
B){5}
C) $\left\{ - \frac { 1 } { 3 } , 1 \right\}$
D){3,1}
E){-3,-1}

Quizplus · Accepted Answer

Using the logarithmic property $\log_a b + \log_a c = \log_a (bc)$, the equation can be rewritten as $\log_4((x-3)(x-1)) = \log_4(8)$. Since the bases are the same, the arguments must be equal: $(x-3)(x-1) = 8$. Expanding and solving the quadratic equation $x^2 - 4x + 3 = 8$ gives $x^2 - 4x - 5 = 0$. Factoring or using the quadratic formula yields $x = 5$ or $x = -1$, but $x = -1$ does not satisfy the original equation because it would result in a logarithm of a negative number. Therefore, the only solution is $x = 5$.

The Solution Set of the Equation log⁡4(x−3)+log⁡4(x−1)=log⁡48\log _ { 4 } ( x - 3 ) + \log _ { 4 } ( x - 1 ) = \log _ { 4 } 8log4​(x−3)+log4​(x−1)=log4​8

The Solution Set of the Equation $\log _ { 4 } ( x - 3 ) + \log _ { 4 } ( x - 1 ) = \log _ { 4 } 8$