Solve the logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic
expressions. Give the exact answer.
-$\log x + \log ( x - 1 ) = \log 30$
A) $\{ 6 \}$
B) $\{ - 5 \}$
C) $\{ 6 , - 5 \}$
D) $\left\{ \frac { 31 } { 2 } \right\}$

Question

Quizplus · Accepted Answer

Using the property of logarithms that $\log a + \log b = \log(ab)$, the equation becomes $\log(x(x - 1)) = \log 30$. Solving $x(x - 1) = 30$ gives $x^2 - x - 30 = 0$. Factoring or using the quadratic formula yields $x = 6$ or $x = -5$. However, $x = -5$ is not in the domain of the original logarithmic expressions because $\log(x)$ and $\log(x - 1)$ are undefined for $x \leq 0$ and $x \leq 1$, respectively. Therefore, the only solution is $x = 6$.

Solve the Logarithmic Equation log⁡x+log⁡(x−1)=log⁡30\log x + \log ( x - 1 ) = \log 30logx+log(x−1)=log30

Solve the Logarithmic Equation $\log x + \log ( x - 1 ) = \log 30$