The frequency of a note on the piano, in Hz, is related to its position on the keyboard by the function f(n) = 440 - 21 where n is the number of keys (both white and black) above or below the note concert A. (Concert A, with a frequency of 440 hertz, is the frequency of the A key above the middle C key on a piano.) Using this function f(n), find the position n (relative to the position of concert A) of the piano key that has a frequency of 55 Hz. (In other words, what number n do you need to input in the function f (n) for f (12) = 55?) Show your steps and explain your solution process. ​

Respuesta :

Step-by-step explanation:

I think you made some typos in the problem definition.

particularly the function definition is wrong.

the correct function to calculate the frequency for every key of a piano keyboard is

f(n) = 440 Ă— 2^((n-49)/12)

now we have a given frequency and do the reverse function on it to find the key number with that frequency.

55 = 440 Ă— 2^((n-49)/12)

55/440 = 2^((n-49)/12)

log2(55/440) = (n-49)/12

12Ă—log2(55/440) = n - 49

n = 12Ă—log2(55/440) + 49 = 12Ă—log2(1/8) + 49 =

= 12Ă—-3 + 49 = -36 + 49 = 13

the frequency of the 13th key is 55 Hz.

f(13) = 55 Hz.

about log2(1/8) :

remember that 1/(x^n) can also be written as x^(-n).

the 1/... operation creates a negative exponent.

8 = 2Âł

so, 1/8 = 2^(-3)

and therefore log2(2^(-3)) = -3