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There are 7 notes of "A" on a standard piano. When the third A is struck, the string associated with the key vibrates 220 times per second. The next A above the given A vibrates twice as fast. An exponential function with a base of two is used to determine the frequency of the 11 notes between the two A's.

Find this function. Hint: The function is of the form:

\[ f(x) = k \cdot 2^{cx} \]

where \( k \) and \( c \) are constants. Also, \( f(0) = 220 \) and \( f(12) = 440 \).

\[ f(x) = \]

Answer :

The exponential function representing the frequencies of the 11 notes between the two A's is: [tex]\(f(x) = 220 \cdot 2^{\frac{\ln(2)}{12} \cdot x}\).[/tex]

To find the exponential function that represents the frequencies of the 11 notes between the two A's, we can use the given information and the hint provided. The function is of the form:

[tex]\[f(x) = k \cdot 2^{cx}\][/tex]

We are given two key pieces of information:

1. When the third A is struck, the string vibrates 220 times per second, so f(0) = 220.

2. The next A above the given A vibrates twice as fast, so f(12) = 440.

Now, we can use these values to solve for the constants k and c.

1. Using f(0) = 220:

[tex]\[220 = k \cdot 2^{c \cdot 0} = k \cdot 2^0 = k \cdot 1 = k\][/tex]

So, we have found that k = 220.

2. Using f(12) = 440:

[tex]\[440 = 220 \cdot 2^{c \cdot 12}\][/tex]

Now, divide both sides by 220 to isolate [tex]\(2^{c \cdot 12}\):[/tex]

[tex]\[2^{c \cdot 12} = \frac{440}{220} = 2\][/tex]

Take the natural logarithm (ln) of both sides to solve for [tex]\(c \cdot 12\)[/tex]:

[tex]\[c \cdot 12 = \ln(2)\][/tex]

Now, divide by 12 to find c:

[tex]\[c = \frac{\ln(2)}{12}\][/tex]

So, the exponential function that represents the frequencies of the 11 notes between the two A's is:

[tex]\[f(x) = 220 \cdot 2^{\frac{\ln(2)}{12} \cdot x}\][/tex]

To know more about exponential function, refer here:

https://brainly.com/question/35259468

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