Solve.
-$\log _{4}(6 \mathrm{x}-7)=\log _{4}(4 \mathrm{x}+2)$
A) $\left\{\frac{9}{2}\right\}$
B) $\{-5\}$
C) $\left\{-\frac{5}{2}\right\}$
D) $\varnothing$

Name: Solve. -$\log _{4}(6 \mathrm{x}-7)=\log _{4}(4 \mathrm{x}+2)$ A) $\left\{\frac{9}{2}\right\}$ B) $\{-5\}$ C) $\left\{-\frac{5}{2}\right\}$ D) $\varnothing$
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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: $6x - 7 = 4x + 2$. Solving for $x$ gives $x = \frac{9}{2}$.

Solve - log⁡4(6x−7)=log⁡4(4x+2)\log _{4}(6 \mathrm{x}-7)=\log _{4}(4 \mathrm{x}+2)log4​(6x−7)=log4​(4x+2) A) {92}\left\{\frac{9}{2}\right\}{29​} B) {−5}\{-5\}{−5} C) {−52}\left\{-\frac{5}{2}\right\}{−25​} D) ∅\varnothing∅

Solve
- $\log _{4}(6 \mathrm{x}-7)=\log _{4}(4 \mathrm{x}+2)$ A) $\left\{\frac{9}{2}\right\}$ B) $\{-5\}$ C) $\left\{-\frac{5}{2}\right\}$ D) $\varnothing$