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A. Simple and Compound Interest

A. Simple and Compound Interest

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CHAPTER 5 Exponential and Logarithmic Functions



Compound Interest

Many financial institutions pay compound interest on deposits they receive, which is

interest paid on previously accumulated interest. The most common compounding periods are yearly, semiannually (two times per year), quarterly (four times per year),

monthly (12 times per year), and daily (365 times per year). Applications of compound

interest typically involve exponential functions. For convenience, consider $1000 in principal, deposited at 8% for 3 yr. The simple interest calculation shows $240 in interest is

earned and there will be $1240 in the account: A ϭ 1000 31 ϩ 10.082132 4 ϭ $1240. If

the interest is compounded each year 1t ϭ 12 instead of once at the start of the 3-yr period,

the interest calculation shows

A1 ϭ 100011 ϩ 0.082 ϭ 1080 in the account at the end of year 1,





A2 ϭ 108011 ϩ 0.082 ϭ 1166.40 in the account at the end of year 2,





A3 ϭ 1166.4011 ϩ 0.082 Ϸ 1259.71 in the account at the end of year 3.

The account has earned an additional $19.71 interest. More importantly, notice

that we’re multiplying by 11 ϩ 0.082 each compounding period, meaning results can

be computed more efficiently by simply applying the factor 11 ϩ 0.082 t to the initial

principal p. For example,

A3 ϭ 100011 ϩ 0.082 3 Ϸ $1259.71.

In general, for interest compounded yearly the accumulated value is

A ϭ p11 ϩ r2 t. Notice that solving this equation for p will tell us the amount we need

A

to deposit now, in order to accumulate A dollars in t years: p ϭ 11 ϩ

r2 t . This is called

the present value equation.

Interest Compounded Annually

If a principal p is deposited at interest rate r and compounded yearly for a period of

t yr, the accumulated value is

A ϭ p11 ϩ r2 t

If an accumulated value A is desired after t yr, and the money is deposited at interest

rate r and compounded yearly, the present value is





EXAMPLE 2







A

11 ϩ r2 t



Finding the Doubling Time for Interest Compounded Yearly

An initial deposit of $1000 is made into an account paying 6% compounded yearly.

How long will it take for the money to double?



Solution







Using the formula for interest compounded yearly we have

A ϭ p11 ϩ r2 t

2000 ϭ 100011 ϩ 0.062 t

2 ϭ 1.06 t

ln 2 ϭ t ln 1.06

ln 2

ϭt

ln 1.06

11.9 Ϸ t



given

substitute 2000 for A, 1000 for p, and 0.06 for r

isolate variable term

apply base-e logarithms; power property

solve for t

approximate form



The money will double in just under 12 yr.

Now try Exercises 17 through 22







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Section 5.6 Applications from Business, Finance, and Science



If interest is compounded monthly (12 times each year), the bank will divide the

interest rate by 12 (the number of compoundings), but then pay you interest 12 times

per year (interest is compounded). The net effect is an increased gain in the interest you

earn, and the final compound interest formula takes this form:

total amount ϭ principal a1 ϩ



1years ϫcompoundings per year2

interest rate

b

compoundings per year



Compounded Interest Formula

If principal p is deposited at interest rate r and compounded n times per year for a

period of t yr, the accumulated value will be:

r nt

A ϭ p a1 ϩ b

n



EXAMPLE 3







Solving an Application of Compound Interest

Macalyn won $150,000 in the Missouri lottery and decides to invest the money for

retirement in 20 yr. Of all the options available here, which one will produce the

most money for retirement?

a. A certificate of deposit paying 5.4% compounded yearly.

b. A money market certificate paying 5.35% compounded semiannually.

c. A bank account paying 5.25% compounded quarterly.

d. A bond issue paying 5.2% compounded daily.



Solution







a. A ϭ $150,000 a1 ϩ



0.054 120ϫ12

b

1



c. A ϭ $150,000 a1 ϩ



Ϸ $429,440.97



0.0535 120ϫ22

b

b. A ϭ $150,000 a1 ϩ

2

Ϸ $431,200.96

A. You’ve just seen how

we can calculate simple

interest and compound

interest



Ϸ $425,729.59

d. A ϭ $150,000 a1 ϩ



0.0525 120ϫ42

b

4

0.052 120ϫ3652

b

365



Ϸ $424,351.12



The best choice is (b), semiannual compounding at 5.35% for 20 yr.

Now try Exercises 23 through 30







B. Interest Compounded Continuously

It seems natural to wonder what happens to the interest accumulation as n (the number

of compounding periods) becomes very large. It appears the interest rate becomes very

small (because we’re dividing it by n), but the exponent becomes very large (since we’re

multiplying it by n). To see the result of this interplay more clearly, it will help to rewrite

the compound interest formula A ϭ p11 ϩ nr 2 nt using the substitution n ϭ xr. This gives

r

1

1

r

n ϭ x , and by direct substitution 1xr for n and x for n 2 we obtain the form

1 x rt

A ϭ p c a1 ϩ b d

x



by regrouping. This allows for a more careful study of the “denominator versus exponent” relationship using 11 ϩ 1x 2 x, the same expression we used in Section 5.2 to define

the number e (also see Section 5.2 Exercise 101). Once again, note what happens as

x S q (meaning the number of compounding periods increase without bound).

x



1



10



100



1000



10,000



100,000



1,000,000



2



2.59374



2.70481



2.71692



2.71815



2.71827



2.71828



x



1

a1 ϩ b

x



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As before, as x S q, 11 ϩ 1x 2 x S e. The net result of this investigation is a formula for interest compounded continuously, derived by replacing 11 ϩ 1x 2 x with the

number e in the formula for compound interest, where

1 x rt

A ϭ p c a1 ϩ b d ϭ pert

x

Interest Compounded Continuously

If a principal p is deposited at interest rate r and compounded continuously for a

period of t years, the accumulated value will be

A ϭ pert

EXAMPLE 4







Solving an Application of Interest Compounded Continuously

Jaimin has $10,000 to invest and wants to have at least $25,000 in the account in

10 yr for his daughter’s college education fund. If the account pays interest

compounded continuously, what interest rate is required?



Solution









In this case, P ϭ $10,000, A ϭ $25,000, and t ϭ 10.



Algebraic Solution



A ϭ pe

25,000 ϭ 10,000e10r

2.5 ϭ e10r

ln 2.5 ϭ 10r ln e

ln 2.5

ϭr

10

0.092 Ϸ r

rt







given

substitute given values

isolate variable term

use natural logs; power property

solve for r (ln e ϭ 1)

approximate form



Jaimin will need an interest rate of

about 9.2% to meet his goal.

B. You’ve just seen how

we can calculate interest

compounded continuously



Graphical Solution



30,000

Using Y1 ϭ 10,000e10X and

Y2 ϭ 25,000, we look for

their point of intersection.

For the window size, since

0

0.12

25,000 is the goal,

y ʦ 30, 30,000 4 seems

reasonable for y. Although

12% interest 1x ϭ 0.122

0

is too good to be true,

x ʦ [0, 0.12] will create a nice frame for the x-values. The point

of intersection shows an interest rate of about 9.2% is required.



Now try Exercises 31 through 40







C. Applications Involving Annuities and Amortization

Our previous calculations for simple and compound interest involved a single (lump)

deposit (the principal) that accumulated interest over time. Many savings and investment plans involve a regular schedule of deposits (monthly, quarterly, or annual deposits) over the life of the investment. Such an investment plan is called an annuity.

Suppose that for 4 yr, $100 is deposited annually into an account paying 8% compounded yearly. Using the compound interest formula we can track the accumulated

value A in the account:

A ϭ 100 ϩ 10011.082 1 ϩ 10011.082 2 ϩ 10011.082 3



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To develop an annuity formula, we multiply the annuity equation by 1.08, then

subtract the original equation. This leaves only the first and last terms, since the other

(interior) terms add to zero:



WORTHY OF NOTE

It is often assumed that the first

payment into an annuity is made at

the end of a compounding period,

and hence earns no interest. This is

why the first $100 deposit is not

multiplied by the interest factor.

These terms are actually the terms

of a geometric sequence, which

we will study later in Section 9.3.



1.08A ϭ 10011.082 ϩ 10011.082 2 ϩ 10011.082 3 ϩ 10011.082 4

2















ϪA ϭ Ϫ 3100 ϩ 10011.082 ϩ 10011.082 ϩ 10011.082 4

1



3



1.08A Ϫ A ϭ 10011.082 Ϫ 100

4



0.08A ϭ 100 3 11.082 4 Ϫ 1 4



multiply by 1.08

original equation

subtract (“interior

terms” sum to zero)

factor out 100



100 3 11.082 Ϫ 1 4

4







solve for A



0.08



This result can be generalized for any periodic payment P, interest rate r, number

of compounding periods n, and number of years t. This would give

r nt

P c a1 ϩ b Ϫ 1 d

n



r

n

The formula can be made less formidable using R ϭ nr , where R is the interest rate

per compounding period.

Accumulated Value of an Annuity

If a periodic payment P is deposited n times per year at an annual interest rate r with

interest compounded n times per year for t years, the accumulated value is given by





P

r

3 11 ϩ R2 nt Ϫ 1 4, where R ϭ

n

R



This is also referred to as the future value of the account.



EXAMPLE 5







Solving an Application of Annuities

Since he was a young child, Fitisemanu’s parents have been depositing $50 each

month into an annuity that pays 6% annually and is compounded monthly. If the

account is now worth $9875, how long has it been open?



Solution







Algebraic Solution







In this case P ϭ 50, r ϭ 0.06, n ϭ 12, R ϭ 0.005, and A ϭ 9875.

P

3 11 ϩ R2 nt Ϫ 1 4

R

50

9875 ϭ

3 11.0052 11221t2 Ϫ 1 4

0.005

1.9875 ϭ 1.00512t

ln11.98752 ϭ 12t1ln 1.0052

ln11.98752

ϭt

12 ln11.0052

11.5 Ϸ t





future value formula



substitute given values

simplify and isolate variable term

apply base-e logarithms; power property

solve for t (exact form)

approximate form



The account has been open approximately 11.5 yr.



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