Solve.
-$\log _{2}(4 x+8)=\log _{2}(4 x+5)$
A) $\{3\}$
B) $\varnothing$
C) $\{0\}$
D) $\left\{\frac{13}{3}\right\}$

Name: Solve. -$\log _{2}(4 x+8)=\log _{2}(4 x+5)$ A) $\{3\}$ B) $\varnothing$ C) $\{0\}$ D) $\left\{\frac{13}{3}\right\}$
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Question

Quizplus · Accepted Answer

Since the logarithms are equal and have the same base, the arguments must be equal. Setting $4x + 8 = 4x + 5$, we see that no value of $x$ satisfies this equation, leading to no solution.

Solve - log⁡2(4x+8)=log⁡2(4x+5)\log _{2}(4 x+8)=\log _{2}(4 x+5)log2​(4x+8)=log2​(4x+5) A) {3}\{3\}{3} B) ∅\varnothing∅ C) {0}\{0\}{0} D) {133}\left\{\frac{13}{3}\right\}{313​}

Solve
- $\log _{2}(4 x+8)=\log _{2}(4 x+5)$ A) $\{3\}$ B) $\varnothing$ C) $\{0\}$ D) $\left\{\frac{13}{3}\right\}$