Solve.
-$\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$
A) $\{-24\}$
B) $\left\{\frac{10}{2}\right\}$
C) $\varnothing$
D) $\{-24,24\}$

Name: Solve. -$\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A) $\{-24\}$ B) $\left\{\frac{10}{2}\right\}$ C) $\varnothing$ D) $\{-24,24\}$
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Question

Quizplus · Accepted Answer

Since the bases are the same, we can set the exponents equal to each other.$$x^{2}+3 x+19=x^{2}+x-29$$$$3 x+19=x-29$$$$2 x=-48$$$$x=-24$$

Solve - log⁡6(x2+3x+19)=log⁡6(x2+x−29)\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)log6​(x2+3x+19)=log6​(x2+x−29) A) {−24}\{-24\}{−24} B) {102}\left\{\frac{10}{2}\right\}{210​} C) ∅\varnothing∅ D) {−24,24}\{-24,24\}{−24,24}

Solve
- $\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A) $\{-24\}$ B) $\left\{\frac{10}{2}\right\}$ C) $\varnothing$ D) $\{-24,24\}$