Solve: $\log _ { 3 } ( x - 5 ) + \log _ { 3 } ( x + 3 ) = \log _ { 3 } 20$
A) no solution
B) {-5}
C) {5}
D) {7}

Question

Quizplus · Accepted Answer

Using the logarithmic property that allows combining logs with the same base on the left side: $\log_{3}(x-5) + \log_{3}(x+3) = \log_{3}((x-5)(x+3))$. This simplifies to $\log_{3}((x-5)(x+3)) = \log_{3}20$. Since the bases are the same, we can set the arguments equal to each other: $(x-5)(x+3) = 20$. Expanding and solving the quadratic equation gives $x^2 - 2x - 15 = 20$, which simplifies to $x^2 - 2x - 35 = 0$. Factoring or using the quadratic formula yields $x = 7$ or $x = -5$. However, substituting $x = -5$ back into the original equation gives a log of a negative number, which is undefined. Therefore, the only solution is $x = 7$.

Solve log⁡3(x−5)+log⁡3(x+3)=log⁡320\log _ { 3 } ( x - 5 ) + \log _ { 3 } ( x + 3 ) = \log _ { 3 } 20log3​(x−5)+log3​(x+3)=log3​20

Solve $\log _ { 3 } ( x - 5 ) + \log _ { 3 } ( x + 3 ) = \log _ { 3 } 20$