Use the laws of logarithms to expand the expression. $\ln \left( \frac { x + 5 } { x - 6 } \right) ^ { 1 / 2 }$
A) $\frac { 1 } { 2 } \ln \left( \frac { x + 5 } { x - 6 } \right)$
B) $\frac { 1 } { 2 } \ln [ ( x + 5 ) ( x - 6 ) ]$
C) $\frac { 1 } { 2 } \ln \left( x - \frac { 5 } { 6 } \right)$
D) $\frac { 1 } { 2 } \ln ( x + 5 ) - \frac { 1 } { 2 } \ln ( x - 6 )$

Question

Quizplus · Accepted Answer

We use the rule $(\ln a)^n = \ln a^n$ to bring the $1/2$ exponent outside the logarithm:
\begin{align*}
\ln\left(\frac{x+5}{x-6}ight)^{1/2} &= \frac12 \ln\left(\frac{x+5}{x-6}ight)
\end{align*}
Then we use the property $\ln(ab) = \ln a + \ln b$:
\begin{align*}
\frac12\ln\left(\frac{x+5}{x-6}ight) &= \frac12\big[\ln(x+5) - \ln(x-6)\big]\
&= \frac12 \ln (x+5) - \frac12\ln(x-6)
\end{align*} This matches choice (D).

Use the Laws of Logarithms to Expand the Expression ln⁡(x+5x−6)1/2\ln \left( \frac { x + 5 } { x - 6 } \right) ^ { 1 / 2 }ln(x−6x+5​)1/2

Use the Laws of Logarithms to Expand the Expression $\ln \left( \frac { x + 5 } { x - 6 } \right) ^ { 1 / 2 }$