Chapter 12 - Exponential Functions and Logarithmic Functions - 12.7 Applications of Exponential Functions and Logarithmic Functions - 12.7 Exercise Set - Page 833: 9

a. The 49th key of the keyboard has a frequency of 440 Hz. b. It takes 12 keys to double the frequency.

a. Consider the model $f\left( n \right)=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}}$ Here, $f\left( n \right)$ is the frequency in hertz (Hz) of the $n\text{th}$ key on a 88-key piano. Replace $f\left( n \right)$ with 440 and solve for $n$: $\begin{align} & f\left( n \right)=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}} \\ & 440=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}} \end{align}$ $\begin{align} & \frac{440}{27.5}=\frac{27.5{{\left( \sqrt[12]{2} \right)}^{n-1}}}{27.5} \\ & 16={{\left( \sqrt[12]{2} \right)}^{n-1}} \end{align}$ Take common logarithm on both the sides: $\begin{align} & \log 16=\log {{\left( \sqrt[12]{2} \right)}^{n-1}} \\ & \log 16=\log {{2}^{\frac{n-1}{12}}} \\ \end{align}$ Apply the power rule of logarithms $\log {{m}^{n}}=n\log m$ in $\log {{2}^{\frac{n-1}{12}}}$: $\begin{align} & \log 16=\log {{2}^{\frac{n-1}{12}}} \\ & \log 16=\frac{\left( n-1 \right)}{12}\log 2 \\ \end{align}$ Divide by $\log 2$ on both sides: $\begin{align} & \frac{\log 16}{\log 2}=\frac{\frac{\left( n-1 \right)}{12}\log 2}{\log 2} \\ & \frac{\log 16}{\log 2}=\frac{n-1}{12} \end{align}$ Using calculator, $\log 16\approx 1.204$ and $\log 2\approx 0.3010$. Therefore, $\begin{align} & \frac{\log 16}{\log 2}=\frac{n-1}{12} \\ & \frac{1.204}{0.3010}=\frac{n-1}{12} \\ & \left( 4\times 12 \right)+1=n \\ & 49=n \end{align}$ b. First calculate the frequency of the first key. $\begin{align} & f\left( n \right)=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}} \\ & f\left( 1 \right)=27.5{{\left( \sqrt[12]{2} \right)}^{1-1}} \\ & f\left( 1 \right)=27.5{{\left( \sqrt[12]{2} \right)}^{0}} \\ & f\left( 1 \right)=27.5 \\ \end{align}$ Now, if $27.5$is the frequency of the first key, $2\left( 27.5 \right)=55$ is the doubled frequency. Therefore, replace $f\left( n \right)$ with 55 in $f\left( n \right)=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}}$ and then solve for $n$: $\begin{align} & f\left( n \right)=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}} \\ & 55=27.5{{\left( \sqrt[12]{2} \right)}^{n-1}} \end{align}$ Divide by $27.5$ on both sides: $\begin{align} & \frac{55}{27.5}=\frac{27.5{{\left( \sqrt[12]{2} \right)}^{n-1}}}{27.5} \\ & 2={{\left( \sqrt[12]{2} \right)}^{n-1}} \end{align}$ Take the common logarithm on both the sides: $\begin{align} & \log 2=\log {{\left( \sqrt[12]{2} \right)}^{n-1}} \\ & \log 2=\log {{2}^{\frac{n-1}{12}}} \\ \end{align}$ Apply the power rule of logarithms $\log {{m}^{n}}=n\log m$ in $\log {{2}^{\frac{n-1}{12}}}$: $\begin{align} & \log 2=\log {{2}^{\frac{n-1}{12}}} \\ & \log 2=\left( \frac{n-1}{12} \right)\log 2 \\ \end{align}$ Divide by $\log 2$ on both the sides: $\begin{align} & \log 2=\left( \frac{n-1}{12} \right)\log 2 \\ & \frac{\log 2}{\log 2}=\frac{\left( \frac{n-1}{12} \right)\log 2}{\log 2} \\ & 1=\left( \frac{n-1}{12} \right) \end{align}$ Solve further, $\begin{align} & 1=\left( \frac{n-1}{12} \right) \\ & \left( 1\times 12 \right)+1=n \\ & n=13 \end{align}$ It means that the 13th key has a frequency of 55 Hz which is double of the frequency of the first key. There are 12 keys between the 1st and 13th key. So, it takes 12 keys to double the frequency