Chapter 4, Problem 46RE

To determine

To Expand: The Logarithm Expression

  $\log \left(x\sqrt{x^2+1}\right)$

Answer to Problem 46RE

  ${\log }_{2}x+\frac{1}{2}{\log }_2\left(x^2+1\right)$

Explanation of Solution

Given:To Expand the logarithm expression

  $\log \left(x\sqrt{x^2+1}\right)$

FORMULA USED:

  $\begin{array}{l}\text{properties of logarithms:}\\ \log a^x=x\log a\\ {\log }_{a}xy={\log }_{a}x+{\log }_{a}y\end{array}$

Given: $\log \left(x\sqrt{x^2+1}\right)$

Using the property of the logarithms:

  $\begin{array}{l}\log \left(x\sqrt{x^2+1}\right)\\ =\log x+\log \left(\sqrt{x^2+1}\right)\end{array}$

  $\begin{array}{l}={\log }_{2}x+{\log }_2\left(\sqrt{x^2+1}\right)\\ ={\log }_{2}x+{\log }_2{\left(x^2+1\right)}^{\frac{1}{2}}\\ ={\log }_{2}x+\frac{1}{2}{\log }_2\left(x^2+1\right)\\ i.e\\ \log \left(x\sqrt{x^2+1}\right)={\log }_{2}x+\frac{1}{2}{\log }_2\left(x^2+1\right)\end{array}$

Hence, $\log \left(x\sqrt{x^2+1}\right)={\log }_{2}x+\frac{1}{2}{\log }_2\left(x^2+1\right)$