Notes - Exponential and Logarithmic Functions Exponential

Notes - Exponential and Logarithmic Functions
Exponential Functions
Def: A function whose equation contains a variable in the exponent is called an exponential function.
f(x) = bx or y = bx
b is called a base and is a positive constant other than 1.
Ex1
The exponential function f(x) = 13.49(0.967)x - 1 describes the number of O-rings expected to fail, f(x),
when the temperature is xoF. If the temperature is 31oF, how many O-rings are expected to fail?
f(31) = 13.49(0.967)31 – 1
f(31) ≈ 4
You Try: How many O-rings are expected to fail at 60oF?
f(60) = 13.49(0.967)60 – 1
f(60) ≈ 1
Ex2
The function below describes the number of people in the audience, f(x), who have heard a rumor x
minutes after 8:00. Evaluate f(10) and describe what this means in terms of the problem?
!""
f(x) = !!!""(!.!")!
Since there is so much going on in the denominator of this function, I must
!""
f(10) = !!!""(!.!")!"
put extra parentheses around it. It would look like this in the calculator:
400/(1 + 399(0.67)^10)
f(10) ≈ 48 → approximately 48 people heard the rumor by 8:10.
You Try - Evaluate f(20) and describe what this means.
!""
f(20) = !!!""(!.!")!"
f(20) ≈ 353 → approximately 353 people heard the rumor by 8:20.
Def:
The number e is called the natural base and is approximately = 2.72...
Just like π is called pi and is approximately = 3.14...
Ex3
push 2nd ln to get e in the calculator.
The function f(x) = 6e0.013x describes world population, f(x), in billions, x years after 2000. Use the
function to find world population in 2050.
0.013*50
f(50) = 6e
f(50) = 11.5 billion
Interest formulas are exponential functions:
***compounding interest: A = P(1 + r/n)nt
***continuous compounding: A = Pert
A = balance (what you will make)
P = principal (what you are investing)
r = interest rate (in decimal form)
t = time (years)
n = number of compounding
I plug in 50 instead of 2050 because the problem indicates that x
is years after 2000. The year 2050 is 50 years after 2000, so I
must plug in 50 for x.
Compounded:
monthly: n = 12
yearly: n = 1
quarterly: n = 4
semiannually: n = 2
Ex4
You want to invest $8000 for 6 years and you have a choice between two accounts. The first pays 7%
per year, compounded monthly. The second pays 6.85% per year, compounded continuously. Which is
the better investment?
I have to use both formulas because I must choose the best account.
A = P(1 + r/n)nt
A = what I’m finding
P = 8000
r = 0.07 → 7% converts to 0.07
n = 12 → compounded monthly, n = 12
t=6
0.07
A = 8000(1 +
A = 12,160.84
A = Pert
A = what I’m finding
P = 8000
r = 0.0685 → 6.85% converts to 0.0685
t=6
A = 8000e(0.0685*6)
A = 12,066.60
(12*6)
/12)
The better investment is the account that is compounded monthly.
You Try
A sum of $10,000 is invested at an annual rate of 8%. Find the balance in the account after 5 years
subject to quarterly compounding.
P = 10000
r = 0.08
A = 10000(1 + 0.08/4)(4*5)
A = 14,859.47
t=5
n=4
Logarithmic Functions
Def: The inverse function of the exponential function is called the logarithmic function.
y = logbx is equivalent to by = x
Ex5
The percentage of adult height attained by a boy who is x years old can be modeled by:
f(x) = 29 + 48.8 log (x + 1)
where x represents the boy's age and f(x) represents the percentage of his adult height. Approximately
what percent of his adult height is a boy at age 8?
f(8) = 29 + 48.8 log (8 + 1)
f(8) ≈ 76%
Approximately what percent of his adult height is a boy at age 10?
f(10) = 29 + 48.8 log (10 + 1)
f(10) ≈ 80%
Def:
The inverse of the natural base e is called the natural log, ln.
Ex6
The annual amount that we spend to attend sporting events can be modeled by f(x) = 2.05 + 1.3 ln x
where x represents the number of years after 1984 and f(x) represents the total amount in billions of
dollars. About how much money was spent in 2000?
f(16) = 2.05 + 1.3 ln 16
I plug in 16 because the problem indicates that x
is the years after 1984. The year 2000 is 16 years
f(16) = 5.65 billion
after 1984.
Estimate how much will be spent in 2015.
f(31) = 2.05 + 1.3 ln 31 = 6.51 billion

Download Report