Chapter 4.4, Problem 38E

To determine

To calculate: The expression to use the laws of logarithm, $In\frac{3x^2}{{\left(x+1\right)}^{10}}$

Answer to Problem 38E

The expended form is ${\log }_{e}3-8{\log }_{e}x$ ,

Explanation of Solution

Given information:

Expend the expression, $In\frac{3x^2}{{\left(x+1\right)}^{10}}$

Formula used:

For change the base rule,

  $\begin{array}{l}{\log }_{a}b=\frac{\log b}{\log a},\\ {\log }_a{b}^x=x{\log }_{a}b\end{array}$

Calculation :

Consider the expression, $In\frac{3x^2}{{\left(x+1\right)}^{10}}$

Rewrite the expression,

  $In\frac{3x^2}{{\left(x+1\right)}^{10}}={\log }_{e}3x^2-{\log }_e{\left(x+1\right)}^{10}$

Apply the log base rule: , ${\log }_a\left(x\right)=\frac{{\log }_b\left(x\right)}{{\log }_b\left(a\right)},{\log }_a{b}^x=x{\log }_{a}b$

  ${\log }_{e}3x^2-{\log }_e{\left(x+1\right)}^{10}={\log }_{e}3+2{\log }_{e}x-10{\log }_{e}x-10{\log }_{e}1$

Here , ${\log }_{e}1=0$ , ${\log }_{e}3+2{\log }_{e}x-10{\log }_{e}x-10{\log }_{e}1={\log }_{e}3-8{\log }_{e}x$

The above expression can’t be simplified further.

Thus the simplified form of the expression is ${\log }_{e}3-8{\log }_{e}x$ .