Exponentiation and Applications

**Exponentiation** is a mathematical operation, written
as \(b^n\), involving two numbers, the
base \(b\) and the exponent or power
\(n\). When \(n\) is a positive integer, \(b^n\) represents the \(n\) times repeated multiplication of the
base \(b\), that is

\[b^{n}=\underbrace{b\times \cdots \times b}_{n},\]

and is pronounced as “\(b\) raised to the power of \(n\)”.

In Excel, \(b^n\) is calculated by
the formula `=b^n`

.

Make a table (see page 597 in the textbook) of the frequency of 2
octaves starting from middle C with a frequency of 260 cps. Use
`=260*1.05946^N`

for \(N\)
half-steps over middle C. Your table may look like this:

half steps above middle C | Note | Frequency |
---|---|---|

0 | C | 260 |

1 | C# | 275.5 |

2 | D | 291.8 |

**Solution:** In a new worksheet, enter “half steps
above middle C” in the cell `A1`

, “Note” in the cell
`B1`

, and “Frequency” in the cell `C1`

.

Enter 0, 1, 2 in the cells `A2`

, `A3`

and
`A4`

. Then select `A2`

to `A4`

and use
autofill to generate numbers up to 24.

Enter 260 in the cell `C2`

and the formula
`=260*1.05946^(N)`

in `C3`

. Then use autofill to
find frequencies for all notes within 2 octave above the middle C.

Enter the name of the notes in colum `B`

and you will get
the table.

Make a table to represent the growth of your initial investment of
$500 at 4% interest compounded annually for 10 years. Use
`=500*(1+0.04)^N`

for a formula. Your table may look like
this:

Years | Balance in $ |
---|---|

0 | 500.00 |

1 | 520.00 |

2 | 540.80 |

How long would it take for your investment to grow to $800 or more?

**Solution:** In a new worksheet, enter “Years” in the
cell `A1`

and “Balance(in $)”, in the cell
`B1`

.

Enter 0, 1, 2 in the cells `A2`

, `A3`

and
`A4`

. Then select `A2`

to `A4`

and use
autofill to generate numbers up to 10.

Enter 500 in the cell `B2`

and the formula
`=500*(1+0.04)^N`

in `B3`

. Then use autofill to
create formulas for each \(n\)-th
year.

For the second part of the question, generate more rows until you see the balance near $800. You will see after 12 years, the balance will be over $800.

**Remark:** If one would like to apply for a loan $L,
with the annual interests \(r\)
compounded \(n\) times for t years,
then the period payment is given by the formula
`=L(1+r/n)^(-n*t)`

.

Make a table of the frequency of 2 octaves starting from one octave below middle C with a frequency of 260 cps to one octaves above middle C. Use

`=260*1.05946^N`

for \(N\) half-steps over middle C.**Hint:**Put 0 for middle C in A13. Then -1 in A12, -2 in A11 and autofill (upward to A1). Put 1 in A14, 2 in A15 and then autofill (downward) to A25. Then generate the frequency using the formula`=260*1.05946^N`

.(

**Optional**) Make a table to represent the growth of your initial investment of $1000 at 4% interest compounded monthly for 20 years. Use`=500*(1+0.04/12)^(12*N)`

for a formula, where \(N\) is the number of years.How long would it take for your investment to grow to $10,000?