Solve.
-$\log _{4} x^{2}=\log _{4}(2 x+15)$
A) $\varnothing$
B) $\{-3\}$
C) $\{5,-3\}$
D) $\{5\}$

Name: Solve. -$\log _{4} x^{2}=\log _{4}(2 x+15)$ A) $\varnothing$ B) $\{-3\}$ C) $\{5,-3\}$ D) $\{5\}$
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Question

Quizplus · Accepted Answer

Since the bases of the logarithms are the same, we can set the arguments equal to each other: $x^2 = 2x + 15$. Solving the quadratic equation $x^2 - 2x - 15 = 0$ by factoring or using the quadratic formula gives $x = 5$ and $x = -3$.

Solve - log⁡4x2=log⁡4(2x+15)\log _{4} x^{2}=\log _{4}(2 x+15)log4​x2=log4​(2x+15) A) ∅\varnothing∅ B) {−3}\{-3\}{−3} C) {5,−3}\{5,-3\}{5,−3} D) {5}\{5\}{5}

Solve
- $\log _{4} x^{2}=\log _{4}(2 x+15)$ A) $\varnothing$ B) $\{-3\}$ C) $\{5,-3\}$ D) $\{5\}$