Chapter 4.7, Problem 2MRPS

Solution

a.

To write: The given expression as a single logarithm.

  $1200{\log }_2\frac{E_4}{C_4}-1200{\log }_2\frac{G_4}{E_4}=1200\left({\log }_2\frac{{\left(E_4\right)}^2}{C_4{G}_4}\right)$

Given:

  $1200{\log }_2\frac{E_4}{C_4}-1200{\log }_2\frac{G_4}{E_4}$

Calculation:

Consider,

  $1200{\log }_2\frac{E_4}{C_4}-1200{\log }_2\frac{G_4}{E_4}$

On simplification, it follows that

  $\begin{array}{c}1200{\log }_2\frac{E_4}{C_4}-1200{\log }_2\frac{G_4}{E_4}=1200\left({\log }_2\frac{E_4}{C_4}-{\log }_2\frac{G_4}{E_4}\right)\\ =1200\left({\log }_2\frac{\frac{E_4}{C_4}}{\frac{G_4}{E_4}}\right)\left[\because {\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)\right]\\ =1200\left({\log }_2\frac{{\left(E_4\right)}^2}{C_4{G}_4}\right)\end{array}$

b.

To evaluate: The expression in part (a) for the given values.

The value of the expression is 70.6724.

Given:

C4 has a frequency of 264 hertz, E4 has a frequency of 330 hertz, and G4 has a frequency of 396 hertz.

Calculation:

By substituting $C_4=264,E_4=330,G_4=396$ in $1200\left({\log }_2\frac{{\left(E_4\right)}^2}{C_4{G}_4}\right)$ , it follows that

  $\begin{array}{c}1200\left({\log }_2\frac{{\left(E_4\right)}^2}{C_4{G}_4}\right)=1200\left({\log }_2\frac{{\left(330\right)}^2}{264\times 396}\right)\\ =1200{\log }_2\left(1.04166\right)\\ =70.6724\end{array}$

Hence, the value of the expression is 70.6724.