Musical notes and exponentials and logarithms This is an exercise from Calculus by Gilbert Strang (under Chapter 6, Exponentials and Logarithms, Section 6.1, An Overview).

The frequency of A above middle C is 440/second. The frequency of the next higher A is _____. Since $2^{7/12} \approx 1.5$, the note with frequency 660/sec is _____.

After reading it for the first time, I did not understand anything. But later I saw the note and frequency and thought the A and the C might be musical notes. I still could not figure it out. I thought about exponentials and logarithms, $10^{440}$ or $\log 440$? They did not make sense, so I searched for their frequencies online. The first blank is 880/second. The second blank is $E_5$. But I still do not understand. What do they have to do with exponentials and logarithms?
 A: If $440/$second is the frequency of the middle A, then the next A is $880/$second. Also, the twelve steps from A to A are approximately multiples of $2^{1/12}$ because we know that frequencies follow a $G.P$ and that there is a $2$ ratio of the octave. So, then seven steps multiplies it by $2^{7/12}$ which is approximately $1.5$ to give $1.5\times 440= 660$. The seventh note from A, as we know is E. Hope it helps.
A: The frequencies in the tempered scale follow a geometric progression rather than an arithmetic one. The ratio between two notes is constant, not the difference.
The ratio of the octave is exactly $2$. (Hence $440\to880$). As there are twelve intervals, the ratio between two notes is such that
$$r^{12}=2$$ or $$r=2^{1/12}.$$
(In terms of logarithms, $\log_2r=\dfrac1{12}$ and the logarithm of the frequencies follows an arithmetic progression.)
As $r^7=2^{7/12}\approx1.5$, the the requested note is seven intervals after $A$, i.e. $E$.

A: The frequency doubles every octave, which consists of 12 half-notes.
Thus, for every half step you take, you multiply (going up) or divide (going down) the frequency by $\sqrt[12]{2}$.
A: (The answers for the blanks are bolded.)
The answer to the first blank is $880$/second, since the frequency doubles every time you go to the next higher note with the same name. 
For the second blank, observe that every time you go up one half-tone, or $\frac 1{12}$ of the distance between the first $A$ the $A$ higher than that, the frequency gets multiplied by $2^{1/12}$. Thus, this is an exponential function. Since $2^{7/12}$ is approximately $1.5$, the note has frequency of $440(1.5)$, or $660$. This note is also $7$ half-tones higher than the original $A$, so that note is $E$, although it would be slightly off-tune because $2^{7/12}$ is (to $4$ decimal places) $1.4983$.
