Expand the logarithmic expression. $\ln \left( x \sqrt { x ^ { 2 } + 1 } \right)$
A) $\ln x + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right)$

B) $\ln x + 2 \ln \left( x ^ { 2 } + 1 \right)$

C) $2 \ln x - \ln \left( x ^ { 2 } + 1 \right)$

D) $\ln x + \ln \left( x ^ { 2 } - 1 \right)$

E) $\frac { 1 } { 2 } \ln x + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right)$

Question

Expand the logarithmic expression. $\ln \left( x \sqrt { x ^ { 2 } + 1 } \right)$
A) $\ln x + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right)$

B) $\ln x + 2 \ln \left( x ^ { 2 } + 1 \right)$

C) $2 \ln x - \ln \left( x ^ { 2 } + 1 \right)$

D) $\ln x + \ln \left( x ^ { 2 } - 1 \right)$

E) $\frac { 1 } { 2 } \ln x + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right)$

Quizplus · Accepted Answer

We can use the properties of logarithms to expand the expression:
\begin{align*}
\ln \left(x\sqrt{x^2+1}ight) &= \ln x + \ln\left(\sqrt{x^2+1}ight) \
&= \ln x + \frac{1}{2}\ln(x^2+1)
\end{align*}
Therefore, the answer is choice A.

Expand the Logarithmic Expression ln⁡(xx2+1)\ln \left( x \sqrt { x ^ { 2 } + 1 } \right)ln(xx2+1​)

Expand the Logarithmic Expression $\ln \left( x \sqrt { x ^ { 2 } + 1 } \right)$