Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
Section 4.1 (cont.) & Section 4.2:
Logarithmic Functions
April 14th, 2014
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
We will first finish Section 4.1 with a discussion of compound
interest. This is an important application of exponential
functions to the worlds of business and finance.
We will then begin our discussion of logarithmic functions. As
we will see, logarithms are the inverses of exponentials, and so
they play a similar role in applications.
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
Suppose P dollars (principal) are invested at an annual
interest rate r and the accumulated value (called future
value) in the account after t years B(t) dollars. If the interest
is compounded k times per year, then
r kt
.
B(t) = P 1 +
k
If the interest is compounded continuously,
B(t) = Pe rt .
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
Example
Suppose $4000 is invested at an annual interest rate of 6.5%.
Compute the balance after 12 years if the interest is
compounded
a. quarterly
b. daily
c. continuously
a. Since $4000 is invested at an annual interest rate of 6.5%,
we have P = 4000 and r = 0.065. By quarterly, we mean
that the interest is compounded 4 times per year, i.e.
k = 4. Therefore the balance after 12 years is
0.065 4(12)
B(12) = 4000 1 +
≈ $8671.35
4
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
b. Here the interest is compounded daily, and so k = 365.
Therefore
0.065 365(12)
≈ $8725.28
B(12) = 4000 1 +
365
c. Since the interest is compounded continuously, we must
use the formula B(t) = Pe rt . Notice that there is no value
of k here. All we need is P, r , and t. Thus the balance
after 12 years is
B(12) = 4000e 0.065(12) = $8725.89
Notice that the more often interest is compounded, the higher
the balance is.
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
Example
Suppose $6000 is invested at an annual interest rate of 3%.
Compute the balance after 5 years if the interest is
compounded
a. annually
b. monthly
c. continuously
Here P = 6000 and r = 0.03.
a. Since the interest is compounded annually, k = 1.
Therefore the balance is
0.03 1(5)
B(5) = 6000 1 +
= $6955.64
1
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
b. Here the interest is compounded monthly, and so k = 12.
Therefore the balance is
0.03 12(5)
B(5) = 6000 1 +
= $6969.70
12
c. Since the interest is compounded continuously, the
balance is
B(5) = 6000e 0.03(5) = 6971.01
Lesson 36
Example
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
How much money should be invested today at an annual
interest rate of 8% compounded continuously so that 15 years
from now it will be worth $30,000?
Since the interest is compounded continuously, we use the
formula B(t) = Pe rt . We are asked to find how much money
should be invested today to make $30,000 15 years from now,
i.e. we’re looking for a value of P with B(15) = 30, 000.
Finally, r = 0.08. Now
Pe 0.08(15) = 30, 000 ⇒ P = 30, 000e −0.08(15) ≈ $9035.83
Therefore we should invest $9035.83 now to have $30,000 in
15 years.
Logarithmic Functions
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
If x is a positive number, then the logarithm of x to the base
b (b > 0, b 6= 1), denoted logb x, is the number y such that
b y = x; that is
y = logb x if and only if b y = x for x > 0.
Lesson 36
Compound
Interest
Examples
Example
Compute the following expressions.
Logarithms
Logarithm Rules
Examples
log2 16;
24 = 16
log3 1;
30
log2 18 ;
2−3
⇒
=1
=
⇒
1
8
⇒
log2 16 = 4
log3 1 = 0
log2
1
8
= −3
We write ln for loge .
ln e 2 ;
e2 = e2
⇒
ln e 2 = 2
Lesson 36
Compound
Interest
We now introduce some basic logarithm rules. These will be
basic in solving the problems we will encounter.
Logarithm Rules
Examples
Logarithms
Let b be any logarithmic base (b > 0, b 6= 1). Then
Logarithm Rules
Examples
logb 1 = 0 and logb b = 1
and if u and v are any positive numbers, we have
(1) logb u = logb v if and only if u = v
(2) logb (uv ) = logb u + logb v
(3) logb u r = r logb u for any real number r
(4) logb ( vu ) = logb u − logb v
(5) logb b u = u
Lesson 36
Example
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
Simplify the expression.
We want to write the expression as one where we are applying
log to just single variables.
log6 (x 2 y 5 )
log6 (x 2 y 5 ) = log6 (x 2 ) + log6 (y 5 )
= 2 log6 x + 5 log6 y
rule (2)
rule (3)
log3 (x 5 y −2 )
log3 (x 5 y −2 ) = log3 (x 5 ) + log3 (y −2 )
= 5 log3 x − 2 log3 y
rule (2)
rule (3)
Lesson 36
Compound
Interest
Examples
Logarithms
Logarithm Rules
Examples
ln
√
5
t2 + t
p
5
ln t 2 + t = ln(t 2 + t)1/5
1
= ln(t 2 + t)
rule (3)
5
1
= ln(t(t + 1))
5
1
= [ln t + ln(t + 1)]
rule (2)
5
1
1
= ln t + ln(t + 1)
5
5
Lesson 36
ln( x1 +
1
)
x2
ln
Compound
Interest
1
1
+ 2
x
x
= ln
x +1
x2
= ln(x + 1) − ln(x 2 )
Examples
rule (4)
Logarithms
= ln(x + 1) − 2 ln x
Logarithm Rules
Examples
ln
h
i
√
4
√ x
x 3 1−x 2
rule (3)
√
4
p
√
x
√
ln
= ln( 4 x) − ln(x 3 1 − x 2 )
x3 1 − x2
p
1
= ln x − (ln(x 3 ) + ln 1 − x 2 )
4
1
1
= ln x − 3 ln x − ln[(1 + x)(1 − x)]
4
2
1
1
1
= ln x − 3 ln x − ln(1 + x) − ln(1 − x)
4
2
2