Solve the equation.
-$\log _{7}(6 x-1)+\log _{7} \mathrm{x}=1$
A) $\left\{\frac{7}{6}\right\}$
B) $\varnothing$
C) $\left\{-1, \frac{7}{6}\right\}$
D) $\{-1\}$

Question

Quizplus · Accepted Answer

Using the logarithm property $\log_b(m) + \log_b(n) = \log_b(mn)$, the equation becomes $\log_7((6x-1)x) = 1$. Converting to exponential form gives $7^1 = 6x^2 - x$, leading to the quadratic equation $6x^2 - x - 7 = 0$. Solving this equation yields $x = \frac{7}{6}$ as the only solution that makes the original logarithmic expressions defined.

Solve the Equation log⁡7(6x−1)+log⁡7x=1\log _{7}(6 x-1)+\log _{7} \mathrm{x}=1log7​(6x−1)+log7​x=1 A) {76}\left\{\frac{7}{6}\right\}{67​} B) ∅\varnothing∅ C) {−1,76}\left\{-1, \frac{7}{6}\right\}{−1,67​}

Solve the Equation $\log _{7}(6 x-1)+\log _{7} \mathrm{x}=1$ A) $\left\{\frac{7}{6}\right\}$ B) $\varnothing$ C) $\left\{-1, \frac{7}{6}\right\}$