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
pϭ
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
Aϭ
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
Aϭ
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
Aϭ
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
Aϭ
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.