Chapter 3, Problem 15CT

To determine

To expand the expression

  $\ln \frac{x\sqrt{x+1}}{2e^4}$

Answer to Problem 15CT

  $\ln \frac{x\sqrt{x+1}}{2e^4}=\ln x+\frac{1}{2}\ln \left(x+1\right)-\ln 2-4$

Explanation of Solution

Given:

  $\ln \frac{x\sqrt{x+1}}{2e^4}$

Concept Used:

Power property of logarithmic function: $\ln u^n=n\ln u$

Product property of logarithmic function: $\ln uv=\ln u+\ln v$

Quotient property of logarithmic function: $\ln \frac{u}{v}=\ln u-\ln v$

  $\ln e^n=n$

Calculation:

To expand the expression

  $\ln \frac{x\sqrt{x+1}}{2e^4}$

Simplify this using the properties of logarithmic function $\ln uv=\ln u+\ln v$ and $\ln \frac{u}{v}=\ln u-\ln v$ .

As,

  $\begin{array}{c}\ln \frac{x\sqrt{x+1}}{2e^4}=\ln \left(x\sqrt{x+1}\right)-\ln \left(2e^4\right)\\ =\ln x+\ln \sqrt{x+1}-\left[\ln 2+\ln e^4\right]\\ =\ln x+\ln {\left(x+1\right)}^{1/2}-\ln 2-\ln e^4\end{array}$

Now to further expand use the property ${\log }_a{u}^n=n{\log }_{a}u$ and $\ln e^n=n$ .

As,

  $\begin{array}{c}\ln \frac{x\sqrt{x+1}}{2e^4}=\ln x+\ln {\left(x+1\right)}^{1/2}-\ln 2-\ln e^4\\ =\ln x+\frac{1}{2}\ln \left(x+1\right)-\ln 2-4\end{array}$

Thus,

  $\ln \frac{x\sqrt{x+1}}{2e^4}=\ln x+\frac{1}{2}\ln \left(x+1\right)-\ln 2-4$