Download An Exponential Function with base b is a function of the form: f(x

An Exponential Function with base b is a function of the form:
f (x) = bx , where b > 0, b 6= 1 is a real number.
We know the meaning of br if r is a rational number. What if r is irrational?
What we do is we approximate the value of br by using rational approximate
for r. For example, to approximate 5π , we may approximate it as 53.14 , 53.141 ,
53.1415 , 53.14159 .... In advance mathematics one can define the value of 5π to be
the limit of these approximations. For now, just realize that such approximation
can be used to approximate the value of br for any positive number b and any
real number r.
Since bx can be defined for all real numbers r, the domain of an exponential
function is all real numbers.
The range of bx is all real numbers greater than 0.
Note the difference between an exponential function and a power function.
x2 is a power function. In this function, the exponent is the constant.
2x is an exponential function. In this function, the base is the constant but the
exponent is the variable (input).
An exponential function is always positive. And if in addition 0 < b < 1, f is a
decreasing function. That is f (x) decreases as x increases.
x
1
f (x) =
2
If b > 1, then f is an increasing function.
I.e. f (x) increases in value as x increases.
f (x) = 2x
Note from the property of exponents that
x
1
1
2−x = x =
2
2
In other words, the exponential functions f (x) = bx and g(x) =
tions of each other with respect to the y axis.
1 x
b
are reflec-
Simple Interest
When you borrow money from a bank, you have to pay interest based on a
particular interest rate. The amount you borrow is called the principal. If
the interest is being calculated based on the principal only, we call this simple
interest. The following formula gives the relationship of these quantities:
Let P = principal, the amount borrowed.
Let r = the simple annual interest rate. r the rate at which interest will be paid
if calculated after 1 year. r is usually given as a percentage, but when used in
calculation, r will be written as a decimal.
Let t = the amount of time, in terms of years, passed when interest is being
calculated.
Let I = interest.
Then we have: I = P rt.
Example: You borrowed $3500 for 15 months, for an interest rate of 6%. How
much interest will you have to pay?
15 5
Ans: P = $ 3500, r = 6% = 0.06. t =
= = 1.25
12 4
I = 3500 × 0.06 × 1.25 = 262.5.
You will have to pay $262.5 in interest.
If you borrowed P in principal, on an annual simple interest rate of r, then after
t years, how much total, A, will you have to pay? In interest problems like this,
the total amount at the end of the period, A, is the future value, and P the
present value.
According to the simple interest formula, you will have to pay I = P rt in interest.
Of course, you will still have to pay off the original amouunt, the principle P , so
the total A you will have to pay is:
A = P + P rt = P (1 + rt)
where A is the total you have to pay (or the future value).
Exmaple: You deposited $8000 into an account that pays 3.5% annual simple
interest. After 18 months, how much should you have in your account?
Ans: P = $8000, r = 3.5% = 0.035, t = 18 months = 1.5 years.
A = 8000(1 + 0.035 × 1.5) = 8420
You will have total of $8420 in your account. $420 is the interest.
Compound Interest:
If you deposit money into a bank (and earns an interest), after your interest is
being paid, if you do not take out the money, the next time you earn interest,
you will earn interest on the original amount, P , and the interest from the first
period, too. When you earn interest on the interest that you earned from a
previous period, you are being paid compound interest.
In calculating compound interest, since you will receive interest on interest that
is being paid from the pervious period, how often you are being paid interest
makes a difference.
Example: You invested $4000 into a compound interest account that pays 12%
rate, compounded quarterly. How much will you have at the end of two years?
Ans: By interest being compounded quarterly, this means that the interest you
1
earn is being calculated (and paid) every 3 months (3 months is of a year).
4
Using the same terminology, if we say interest is compounded monthly, that
means interest will be calculated and paid each month; compounded daily means
interest will be calculated and paid each day...etc.
At the end of the first three months, you will be paid interest. The time elasped
1
since you last received interest is 3 months, or t = year.
4
1
A = P (1 + rt) = 4000 1 + 0.12 4 = 4120
At the end of 6 months, you
be paid interest again, this time, you will already
will
1
in your account. This is your new principal. We
have A = 4000 1 + 0.12
4
use this number in the formula to calculate the new balance:
2
1
1
1
A = 4000 1 + 0.12
1 + 0.12
= 4000 1 + 0.12
≈ 4243.60
4
4
4
At the end of 9 months, you will be paid interest again, this time using yet a new
principal:
2 3
1
1
1
A = 4000 1 + 0.12
· 1 + 0.12
= 4000 1 + 0.12
≈ 4370.91
4
4
4
Looking at the pattern, we see that for each additional compounding period, the
power of the expression (1 + rt), increased by one.
At the end of two years, interest would be have been compounded 8 times, the
total amount we will have is:
8
1
≈ 5067.08
A = 4000 1 + 0.12
4
We have the following formula:
If a principal amount P is earning an annual interest rate of r, and interest is
compounded m times a year, then the total at the end of t years will be given
by:
r mt
A=P 1+
m
Example: $20, 000 is being invested into an account that pays r = 8% nominal
interest. How much will be in the account after 5 years if interest is compounded
a. annualy?
daily?
b. semi-annually?
c. quarterly?
d. monthly?
e.
Ans:
r
0.08
=
= 0.08. In 5 years interest
a. If interest is compounded annually, then
m
1
will be compounded 5 times, so mt = 5, using the formula we have:
A = 20, 000 (1 + 0.08)5 ≈ 29386.56
r
r
0.08
= =
= 0.04. In 5
m
2
2
years interest will be compounded 10 times (two times a year times 5 years), so
mt = 10, using the formula again gives:
10
0.08
A = 20, 000 1 +
≈ 29604.89
2
r
r
0.08
=
=
= 0.02. In 5
c. If interest is compounded quarterly, then
m
4
4
years interest will be compounded 20 times (four times a year times 5 years), so
mt = 20, we have:
20
0.08
A = 20, 000 1 +
≈ 29718.95
4
r
0.08
r
=
=
≈ 0.00667. In 5
d. If interest is compounded monthly, then
m
12
12
years interest will be compounded 60 times (twelve times a year times 5 years),
so mt = 60, we have:
60
0.08
A = 20, 000 1 +
≈ 29796.91
12
r
r
0.08
e. If interest is compounded daily, then
=
=
≈ 0.000219. In 5
m
365
365
years interest will be compounded 1825 times (365 times a year times 5 years),
so mt = 1825, we have:
1825
0.08
A = 20, 000 1 +
≈ 29835.19
365
We see that the more often interest is compounded, the more money we get at
the end. What if we compound interest more often, say compounded every hour,
or every minute, or every second? Will we get infinite amount of money, or will
the amount we get approach a particular number?
b. If interest is compounded semi-annually, then
Looking at the formula
r mt
A=P 1+
m
m
r
1
1
If we let x = , then m = xr,
=
= , then the formula becomes:
r
m m/r
x
xrt
x rt
1
1
A=P 1+
=P
1+
x
x
As we compound more often, m keeps on increasing while r remains the same,
so x also keeps on increasing.
x
1
If we look at the expression, 1 +
, and try some large values of x to see
x
what is the value of the expression, we give a table of the value of the expression
for some large values of x:
x
1
x
1+
x
10
1
10
1+
≈ 2.59
10
100
1
100
1+
≈ 2.7048
100
1000
1
1, 000
1+
≈ 2.7169
1000
10000
1
1+
10, 000
≈ 2.7181
10000
100000
1
100, 000
1+
≈ 2.718
100000
1000000
1
1, 000, 000
1+
≈ 2.71828
1000000
x
1
It appears that the value of the expression 1 +
approaches a particular
x
number as x becomes incresingly large. This is indeed the case. It can be shown
that the limit of this expression, when the value of x becomes arbitrarily large (we
say that the value of x increases
x bound, or that x approaches infinity),
without
1
approaches a particular (irrational) real
the value of the expression: 1 +
x
number. This number is very important in mathematics and science, and is
represented by the letter e and has an approximate value of e ≈ 2.718281828.
The exponential function using e as its base is called the natural expenential
function. In other words, the natural exponential function is the function f (x) =
ex
Using e to replace the above expression, we get the formula:
A = P ert
This is the formula used when interest is compounded continuously. In reality,
we can not really compound interest continuously. The formula gives an upperbound on how much interest can be earned regardless of how often interest is
being compounded.
Example:
$20000 is being invested into an account that pays r = 8% nominal interest. How
much will be in the account after 5 years if interest is compounded continuously?
Ans:
P = 20000, r = 0.08, t = 5
We use the formula for continuous compound interest:
A = 20000e0.08(5) ≈ 29836.5
In other words, given the above condition, you can never earn more than $29836.5
no matter how often interest is compounded.
Logarithmic Functions:
An exponential function is a one-to-one function. It has in inverse. However,
the expression for the inverse of an exponential function cannot be solved by any
algebraic means, therefore we do not have an algebraic expression for it. We just
define such a function and study its property knowing that it is the inverse of
bx
Definition:
g(x) = logb x (read “log base b of x”) is the inverse function of
f (x) = bx
logb is the name of the function. x is the argument (input) to logb , and the value
(output) is logb x.
Since the range of an exponential function is all positive real numbers, the domain
of a log function is all positive real numbers.
Because of the fact that logb x is the inverse of bx , we have this by definition:
logb (bx ) = x for all x
blogb x = x for all x > 0
The following two equations are equivalent:
y = bx
logb y = x
The logarithm with base e, which is the inverse function of ex , is given a special
name, the natural logarithm, and written ln. I.e.
loge x = ln x
Example:
Evaluate:
log2 16
Ans: Set log2 16 = y, this is the same as asking:
2y = 16, what is y? From observation we see that y = 4, so
log2 16 = 4
Example:
1
log5
=?
125
Ans: Since 5−3 =
1
,
125
1
= −3
125
Example:
log5
log2 x = 5, what is x?
Ans: 25 = x, so x = 32
logx 100 = −2, what is x?
This is the same as saying, x−2 = 100, so x =
1
10
Because of the fact that logs are inverse functions of the exponential functions,
they have many properties that are similar to that of the exponential functions,
and can be easily proved using the definition:
Properties of Logarithm
For any real number r, any base b > 0, a > 0, any x > 0, y > 0, we have:
logb (xy) = logb x + logb y
x
logb
= logb x − logb y
y
logb xr = r logb x
Example: ln(20) = ln(4 · 5) = ln(4) + ln(5)
Example: log3 (3xy) = log3 (3) + log3 (x) + log3 (y) = 1 + log3 (x) + log3 (y)
2
= ln(2) − ln(5)
Example: ln
5
Example: log4 (x5 ) = 5 log4 (x)
Example: logb (2x3 y 4 ) = logb (2)+logb (x3 )+logb (y 4 ) = logb (2)+3 logb (x)+4 logb (y)
Change of Base Formula:
logb x
loga x =
logb a
log10 15
Example: log2 (15) =
log10 2
ln(3)
Example: log5 (3) =
ln(5)
The change of base formula says that we can easily change from log of one base
to another, so the choice of log of which base to use usually for convenience
only. We like to use natural log for most of our studies because that is the most
conveninent one in math and science.
Also make sure not to misuse the properties of logs, for example, the property of
logarithm does not give us this: log(x + y) = log x + log y
Here’s some example of how to incorrectly use the property of log:
logb (x + 5) = logb (x) + logb (5)
logb (x2 − 3) = logb (x2 ) − logb (3)
(logb x)3 = 3 logb (x)
To solve an equation involving exponential function, one would need to use logarithm, and to solve an equation involving logarithm, one uses exponential functions.
E.g.
Solve 3x = 10
Ans: We take the log of both side
log3 3x = log3 10
Since log and exponential functions are inverse, the left hand side is just x, we
have
x = log3 10
One may approach the same problem by taking the ln of both sides and use
properties of logarithm:
3x = 10
ln(3x ) = ln 10
x ln 3 = ln 10
ln 10
x=
ln 3
E.g. Solve log4 (x + 2) = 3
Ans: We turn this into an equation involving exponential function:
log4 (x + 2) = 3
43 = x + 2
64 = x + 2
x = 62
E.g. Solve log2 (x + 1) + log2 (x + 4) = 2
Ans: We use property of log to combine the left hand side:
log2 (x + 1) + log2 (x + 4) = 2
log2 ((x + 1)(x + 4)) = 2
log2 (x2 + 5x + 4) = 2
22 = x2 + 5x + 4
4 = x2 + 5x + 4
x2 + 5x = 0
x = 0 or x = −5
Notice that x = −5 is an extraneous solution, so our only solution is x = 0
Application to compound interest:
Example: You invested $8000 into an account that pays 5% nominal rate. How
long before you will have $12000 if interest is compounded:
a. quarterly?
b. monthly?
Ans: We are solving for time t in the formula:
r mt
A=P 1+
m
a. We are told that the nominal rate, r, is r = 5% = 0.05. If compounded
quarterly, then m = 4. You started with P = 8000. You want to end up with
A = 12000, so we have the equation:
4t
0.05
12000 = 8000 1 +
4
This is an exponential equation which can be solved using logarithm:
4t
0.05
12000
= 1+
8000
4
4t
3
0.05
= 1+
2
4
"
4t #
3
0.05
ln
= ln 1 +
2
4
0.05
3
ln
= 4t ln 1 +
2
4
ln 32
4t =
ln 1 + 0.05
4
3
ln 2
≈ 8.16
t=
4 ln 1 + 0.05
4
It takes about 8 years and two months for the $8000 to become $12000
Example: You have some money to invest into an account that pays 6% nominal
interest rate compounded monthly. How long would it take you to double your
investment?
Ans: It does not really matter the amount you started with. The time it takes
you to double your investment is dependent only on the interest rate and the
ways interest is compounded. If you started with initial investment P , you want
to end up with future value of A = 2P .
r = 0.06, m = 12, we want to solve for t:
12t
0.06
2P = P 1 +
12
12t
0.06
2= 1+
12
"
12t #
0.06
ln 2 = ln 1 +
12
0.06
ln 2 = 12t ln 1 +
12
ln 2
≈ 11.58
t=
12 ln 1 + 0.06
12
It takes about 11 and a half years to double your investment.
Example:
You want to deposit some money into an account that pays 3% nominal interest
rate compounded monthly. If you want to be able to withdraw $50000 to buy a
car in 5 years, how much must you deposit now?
Ans: We are trying to find P , given A = 50000, r = 0.05, t = 5, m = 12:
12(5)
0.05
50000 = P 1 +
12
50000
P =
≈ 38960.27
0.05 12(5)
1 + 12
Exponential Growth and Decay:
There are many situations where we may use an exponential function as a model.
For example, the growth of a population may be modeled as an exponential
growth, the amound of radioactive material remain after a certain period of time
may be modeled as an exponential decay.
An exponential growth or decay is defined by the equation:
y = y0 ekt
In this equation, t is time, y0 and k are constants. y0 is the initial value (of the
population, or amount of radioactive material present). If k > 0, the equation
represents an exponential growth, and k is the growth constant. If k < 0, then
the equation represents an exponential decay, and k is the decay constant.
Example: A biologist starts with 2000 bacteria in a petri dish. After three hours,
the population has increased to 2, 600. Assuming that the population of the
bacteria increases exponentially, provide an equation that express the number of
bateria as a function of time.
Ans: Since we assume that the rate of growth of the bacteria is exponential, we
use the equation y = y0 ekt , where y is the number of bacteria at time t.
At time t = 0, the initial population size is
y = y0 ek(0) = 2, 000 ⇒ y0 = 2, 000
This is the initial population, y0 .
After t = 2 hours, the population y has grown to 2, 600, we then set up the
equation:
2, 600 = 2000ek(2) ⇒ e2k =
2600
2000
⇒ e2k =
13
10
⇒ 2k = ln (1.3) ⇒
ln (1.3)
≈ 0.1312
2
The equation we can use to model this population growth is:
k=
y = 2000e0.1312t
How long does it take this bacteria culture to double in size from the original
population size?
Ans: To double in size, the bateria will have to increase to 4000, we set y = 4000
to solve the equation:
ln(2)
≈ 5.28
4000 = 2000e0.1312t ⇒ e0.1312t = 2 ⇒ 0.1312t = ln(2) ⇒ t =
0.1312
It takes approximately 5.28 hours for the population to double in size.
This number, the amount of time it takes for a population to double in size in an
exponential growth, is called the doubling time. This time is independent of
the amount of the initial population. If you started with 50 bacteria, it will take
5.28 hours to double to 100 bacteria. If you started with 5000 bacteria, it will
take the same 5.28 hours to double to 10,000 bacteria.
Example: 50 grams of certain type of radioactive material is placed inside a
detector. After 40 days, it is observed that only 47 grams of the radioactive
material is left. Assuming that the radioactive material decays exponentially,
find an equation that describes the amount of material left as a function of time
t (in days).
Ans: We let y = the amount of radioactive material present at any time t. At
t = 0 we have 50 grams, so we have:
y = y0 ek(0) = 50 ⇒ y0 = 50
The initial amount of radioactive material is 50 grams, as we know.
After t = 40 days, y = 47 grams left, we have:
47
ln
47
47
50
47 = 50ek(40) ⇒ e40k =
⇒ 40k = ln
⇒k=
≈ −0.0015469
50
50
40
The function that can be used to model this radioactive decay is:
y = 50e−0.0015469t
How much time t does it take this radioactive material to decrease to half of what
we started?
Ans: We started with 50 grams, to decrease to half is to end up with only 25
grams, we set up the equation:
25 = 50e−0.0015469t ⇒ e−0.0015469t = 0.5 ⇒ −0.0015469t = ln(0.5) ⇒
ln(0.5)
≈ 448
−0.0015469
It takes approximately 448 days for the radioactive material to decay to half of
what it has originally. This number, the amount of time it takes a radioactive
material to decay to half of its original amount, is called the half-life of a radioactive material, and is independent of the amount of radioactive material
we started with.
t=
Example: After the Chernobyl nuclear meltdown, it is estimated that the concentration of cesium-137, a radioactive material with a half-life of 33 years, was 10
times the normal level. How long will it take the concentration level of cesium-137
to return to normal level.
Ans: We first come up with the equation that models the decay of cesium-137.
Since its half-life is 33 years, it means it takes t = 33 years for the amount to
reduce to half of the original content, so we set up the equation:
1
1
ln(0.5)
y0 = y0 ek(33) ⇒ = e33k ⇒ 33k = ln(0.5) ⇒ k =
≈ −0.021
2
2
33
The equation we can use to mode the decay of cesium-137 is:
y = y0 e−0.021t , where t is expressed in years, and y0 is however much of the
material we started with.
Since the amount we started with, y0 , is 10 times the normal level, we need the
amount to drop down to only one-tenth of y0 to be normal, so we set up this
equation to solve for t:
1
1
ln(0.1)
y0 = y0 e−0.021t ⇒
= e−0.021t ⇒ −0.021t = ln(0.1) ⇒ t =
≈ 109.6
10
10
−0.021
It will take about 110 years for the radiation to drop back down to normal level.