Transcript Slide 1

Exponential functions and their graphs
• The exponential function f with base a is denoted by
f(x) a x
where a  0, a  1, and x is any real number.
• Example 1. If you have a dollar on day zero and you double
your money at each future day and halve your money at each
past day, then as a function of days x, the amount of money
you have is f(x) = 2x. The table below lists some values of
the function.
x
–4
–3
–2
–1
0
1
2
3
4
f(x)
1/16
1/8
1/4
1/2
1
2
4
8
16
• Note that f is an increasing function.
Exponential functions and their graphs, continued
• The domain of f(x) = 2x from the previous slide can be
extended to include all real numbers and the resulting
graph is shown below.
Note that the x-axis is a horizontal asymptote.
Exponential functions and their graphs, continued
• Example 2. If you have a dollar on day zero and you
double your money at each past day and halve your money
at each future day, then as a function of days x, the amount
of money you have is g(x) =  12 x . The table below lists
some values of the function.
x
–4
–3
–2
–1
0
1
2
3
4
g(x)
16
8
4
2
1
1/2
1/4
1/8
1/16
• Note that g is a decreasing function.
Exponential functions and their graphs, continued
• The domain of g(x) =  12  from the previous slide can be
extended to include all real numbers and the resulting
graph is shown below.
x
Note that the x-axis is again a horizontal asymptote.
The relation between Example 1 and Example 2
• In Example 1, f(x) = 2x while in Example 2, g(x) =  12  .
x
• Since 21  1 2 , the formula defining g(x) can be rewritten
as g(x) = 2–x. This means that g(x) = f(–x) and the graph
of g is the graph of f reflected in the y-axis. The plot
below shows both the graph of f and the graph of g.
y

1 x
2
y  2x
Comparing graphs of exponential functions
• Consider f(x) = 2x and h(x) = 4x. Their graphs are shown
below. Notice that h increases faster that f. Also, both
graphs have a y-intercept at (0,1). The exponential function
with the larger base a, a > 1, will always increase faster.
y  4x
y  2x
The one-to-one property
• Since an exponential function is always increasing (a > 1)
or always decreasing (0 < a < 1), its graph passes the
horizontal line test and therefore it is a one-to-one function.
• The one-to-one property for exponential functions is:
For a > 0 and a  1, ax = ay if and only if x = y.
• The one-to-one property can be used to solve simple
exponential equations.
• Solve 8 = 2x+1. Rewrite as 23 = 2x+1 and apply the
one-to-one property to obtain 3 = x+1 so x = 2.
Transformation of graphs of exponential functions
• The transformations of function discussed in Chapter 1 can
be applied to exponential functions. We give one example.
• Example. Let f(x) = 3x. Let g(x) = –f(x) +2. The graph of
g can be obtained from the graph of f by a reflection in the
x-axis followed by a vertical shift of 2 units.
y  f(x)
y  2 is hor.asymptoteof g
y  g(x)
Motivating the number e using compound interest
• Suppose we invest $1.00 at 100% interest once a year. At
the end of the year, we have $2.00.
• Suppose we invest $1.00 at 50% interest twice a year. At
the end of the year, we have $2.25.
• Suppose we invest $1.00 at 25% interest four times a year.
At the end of the year, we have $2.44141.
• As the frequency of compounding increases, the balance at
the end of the year approaches $2.71828..., and this
limiting value is referred to as the number e. It is often
convenient to use the irrational number e as the base for an
exponential function. The number e is referred to as the
natural base.
Compound interest
• Suppose a principal P is invested at an annual interest rate r
compounded once a year. Note that r is a decimal; for example,
a 6% interest rate yields r = 0.06. If this is continued for t years:
Year
Balance after each compounding
0
PP
1
P1  P(1 r)
2
P2  P1 (1 r)  P(1 r)(1 r)  P(1 r)2
3
P3  P2 (1 r)  P(1 r)2 (1 r)  P(1 r)3


t
Pt  P(1 r)t
Formulas for compound interest
• To accommodate more frequent (quarterly, monthly,
weekly, etc.) compounding of interest, let n be the number
of compoundings per year and let t be the number of years.
Then the rate per compounding is r/n and the account
balance after t years is
A  P1 

r nt
n
.
• As the number of compoundings per year increases we have
P1  nr   Pert as n  .
nt
rt
• This limiting value, Pe , gives the balance in the account
for continuous compounding with annual interest rate r
after t years.
Example for compound interest
• If $1000 is invested at an annual interest rate of 5%, find the
balance in the account after 10 years if interest is
compounded (a) quarterly, (b) monthly, and (c) continuously.
• (a) We have P = 1000, r = 0.05, n = 4, t = 10
A  10001 

0.05 (4)(10)
4
 $1643.62
• (b) We have P = 1000, r = 0.05, n =12, t = 10

A  10001  0.05
12
(12)(10)
 $1647.01
• (c) We have P = 1000, r = 0.05, t = 10
A  1000e(0.05)(10)  $1648.72
Solving for an unknown interest rate
• Suppose you make three separate deposits of $1000 each
into a savings account, one deposit per year, beginning
today. What annual interest rate r compounded annually
gives a balance of $3300 three years from today?
• If we let x = 1 + r be the unknown, then the balance after
three years is
1000x 3  1000x 2  1000x
• To find x we must solve
1000x 3  1000x 2  1000x  3300
• This means we must find a zero of the polynomial
Q(x)  x3  x 2  x  3.3
• We may use a calculator or Maple to obtain x = 1.0484.
Therefore, the annual interest rate we want is 4.84%.
Radioactive decay
• A 200 microgram sample of carbon-14 decays according to
the formula
Q  200(0.886)t
where t is in thousands of years. When t = 5.727 (that is
5727 years) we have
5.727
Q  200(0.886)
 100.
• We say that the half-life of carbon-14 is 5727 years
because it takes that long for half of a sample to decay.
Logarithm functions and their graphs
• For x > 0, a > 0, and a  1
loga x  y if and onlyif x  a y .
The function given by
f(x) loga x
is called the logarithm function with base a.
• The logarithm function with base 10 is called the common
logarithm function and it is usually denoted simply as
f(x) = log x. On most calculators, the button LOG is used
for the common logarithm.
• Example. log 1,000,000 is the exponent of 10 that gives
1,000,000. Without using a calculator, can you evaluate
log 1,000,000?
Evaluate each of the following:
• log2 64
• log2 0.5
• log3 1
• log3 3
• log 0.01
• log 2 (Hint--use your calculator.)
The graph, domain, and range of the common logarithm
• It follows from the definition of log x that its domain consists
of all positive real numbers. Its range is all real numbers.
x
log x
0.01
-2
0.1
-1
1
0
10
1
100
2
1000
3
• Using Maple or graphing calculator, we can plot the graph of
log x:
Properties of logarithms
1.
2.
3.
4.
loga 1 = 0 because a0 = 1.
loga a = 1 because a1 = a.
loga ax = x and a loga x  x (inverse properties)
If loga x = loga y, then x = y (one-to-one property)
Example. Solve for x: log4 x 2  2
One way : log4 x 2  log416  x 2  16  x  4.
log4 x 2
2nd way : 4
 42  x 2  16  x  4.
Property 3 may be interpreted as follows:
If f(x) a x , thenf 1 (x)  loga x.
Typical logarithm graph
• When a > 1, a typical graph of loga x is shown along with
some of its properties.
Domain: (0, )
Range: (–, )
y  loga x
x-intercept: (1,0)
Increasing
x
One-to-one => inverse exists
y-axis is a vertical asymptote
Continuous
Reflection of graph of y = ax
about the line y = x
Chemical Acidity
• In chemistry, the acidity of a liquid is expressed using pH.
The acidity depends on the hydrogen ion concentration in
the liquid (in moles per liter). This concentration is written
[H+]. The pH is defined as:
pH   log [H  ].
• Problem. A vinegar solution has a pH of 3. Determine the
hydrogen ion concentration.
Solution. Since 3 = – log[H+], we have –3 = log[H+]. This
means that 10-3 = [H+]. The hydrogen ion concentration is
10-3 moles per liter.
Logarithms and orders of magnitude
• We often compare sizes or quantities by computing their
ratios. If A is twice as tall as B, then
Height of A/Height of B = 2.
• If one object is 10 times heavier than another, we say it is
an order of magnitude heavier. If one quantity is two
factors of 10 greater than another, we say it is two orders
of magnitude greater, and so on.
• Example. The value of a dollar is two orders of magnitude
greater than the value of a penny.
$1
 10 2 .
$0.01
We note that the order of magnitude is the logarithm of the
ratio of their values.
Graphs of 10x and log x
y  10x
· (0.3010, 2)
y  log x
(0,1)
(2, 0.3010)
·
(–1, 0.1)
·
(1,0)
· (0.1, –1)
Shifting the graph of log x
• Let f(x) = log x and g(x) = f(x – 1) = log(x – 1). The graph
of g will be the same as the graph of f shifted one unit to the
right. The graph of g has vertical asymptote x = 1.
y  log x
x
y  log (x  1)
x 1
Note: The domain of f is x > 0
while the domain of g is x > 1.
The natural logarithm function
• The function defined by
f(x) loge x  ln x, x  0
is called the natural logarithm function.
• The symbol ln x is read as "the natural log of x" or "el en
of x". Most calculators will have a button LN for the
natural logarithm.
• If x is a power of e, then it is possible to evaluate ln x
without a calculator. For example, ln e2 = 2. Otherwise,
use a calculator. For example, ln 2 = .6931.
Properties of natural logarithms
1.
2.
3.
4.
ln 1 = 0 because e0 = 1.
ln e =1 because e1 = e.
ln ex = x and eln x = x. (inverse properties)
If ln x = ln y, then x = y. (one-to-one property)
Example. Solve for x: ln(x – 1) = –1.
One way : ln(x 1)  ln e1  x 1  e1  x  1  e1.
2nd way : eln(x1)  e1  x 1  e1  x  1  e1.
Property 3 may be interpreted as follows:
If f(x) ex , thenf 1 (x)  ln x.
Converting from logarithmic form to exponential form
Converting from exponential form to logarithmic form.
Change of base
• Let a, b, and x be positivereal numberssuch thata  1 and b  1.
Then loga x can be converted to a different base as follows.
Base b
Base 10
Base e
logb x
log x
ln x
loga x 
loga x 
loga x 
logb a
log a
ln a
• Example. Evaluate log4 25 using both common and natural
logarithms.
log 25 1.39794
log4 25 

 2.3219
log 4 0.60206
ln 25 3.21888
log4 25 

 2.3219
ln 4
1.38629
Properties of logarithms, continued
• Let a be a positive number such that a  1, and let n be a
real number. If u and v are positive real numbers, the
following properties are true.
1. P roduct P ropert y: loga (uv)  loga u  loga v
ln(uv) ln u  ln v
u
2. Quot ientP ropert y: loga  loga u  loga v
v
u
ln  ln u  ln v
v
3. P ower P ropert y:
loga u n  n  loga u
ln u n  n  ln u
Rewriting logarithmic expressions using properties of logs
• Expand the given expression.
 x2 1
log 3 
 x 
 log(x 2  1)  log x3
 log(x 2  1)  3  log x
• Condense the given expression.
2  log(9  x 2 )  (log(3  x )  log(3 x))
 2  log(9  x 2 )  log[(3  x)(3 x)]
 2  log(9  x 2 )  log(9  x 2 )
 log(9  x 2 )
Decibels
• To measure a sound in decibels, the sound’s intensity, I, in
watts/m2 is compared to a standard benchmark sound, I0.
This results in the following definition:
 I 
,
Noise level in decibels  10  log
 I0 
where I0 is defined to be 10-12 watts/m2, roughly the lowest
intensity audible to humans.
• Problem. If a sound doubles in intensity, by how many
units does its decibel rating increase?
 2I 
 I


Difference in decibel ratings  10  log
 10  log

 I0 
 I0
  2I 
 I 

  log    10  log2  3.010 dB.
 10 log


I
I
0
0









Solving exponential and logarithm equations
•
The following strategies are available, but strategies 2 and 3 are
the most important.
1.
Rewrite the original equation in a form that allows the use of the
one-to-one properties of exponential or logarithm functions.
Rewrite an exponential equation in logarithm form and apply
properties of logarithm functions.
Rewrite a logarithm equation in exponential form and apply
properties of exponential functions.
2.
3.
Example for 1. 2x = 64 => 2x = 26 => x = 6
Example for 2. 2x = 64 => x∙log 2 = log 64 => x = log 64/log 2 = 6
alternatively, 2x = 64 => log2 2x = log2 64 => x = log 64/log 2 = 6
Example for 3. log x = –2 => 10log x = 10–2 => x = 10–2
Solving an exponential equation
• Example. Suppose the temperature H, in °F, of a cup of
coffee t hours after it is set out to cool is given by the
equation:
H  70  120(1/4)t .
How long does it take the coffee to cool down to 90°F?
Solution. We must solve the following equation for t:
70  120(1/4)t  90,
120(1/4)t  20, by subtracting
(1/4) t  1/6, by dividing
log(1/4)t  log(1/6), by takinglogs
t  log(1/4)  log(1/6), using a log property
t  log(1/6)/log(1/4)  1.29 hours.
How many years will it take for your salary to double?
• Problem. If you start at $40000, and you are given a 6%
raise each year, how many years must pass before your
salary is at least $80000?
Solution. We must solve 40000(1.06)t = 80000 for t.
Equivalently, we must solve (1.06)t = 2 for t. If we take
the log of both sides of this equation and use the power
property of logarithms, we obtain
t  log 1.06  log 2, or
log 2
0.30103
t

 11.896 years.
log 1.06 0.025306
If you have to wait until the end of the year to actually get
your raise, 12 years must pass.
More on salary doubling
• Another way to solve (1.06)t = 2 for t is as follows.
log1.061.06t  log1.06 2
log 2
t
 11.896 years, using changeof base
log1.06
• Of course, this is the same answer we obtained previously.
Solving a logarithmic equation
• Solve ln x + ln(x – 2) = 1 for x.
lnx(x  2)   1 using product prop.
e ln x(x  2 )   e1
exponentia
te
x(x  2)  e
inverseprop.
x 2  2x  e  0
algebra
x  1 1 e
quadratic formula
x 1 1 e
ot her" solution"is negative
which is impossiblein original
equat ion
Solving another logarithmic equation
• Solve log x  log(x – 2) = 1 for x.
• Hint: Use the quotient property.
• Answer: x = 20/9.
Solving yet another logarithmic equation
• Solve log2 x  log2 (x  2)  log2 (x  6).
• Answer: x = 2 is the only answer.
Solving an exponential equation
(x2 )
• Solve 2
 4x.
• Answers: x = 0, x = 2.
Half-life of carbon-14
• A 200 microgram sample of carbon-14 decays according to
the formula
Q  200(0.886)t
where t is in thousands of years. How long does it take
until only 100 micrograms remains?
100  200(0.886)t must be solvedfor t
1
2
 (0.886) t
divide by 200
log 12  t  log 0.886 takelogs, use power prop.
t  log 0.5 / log 0.886 5.726
Since t is in thousands of years, the half-life is 5726 years.
Continuous compounding
• Suppose you have two bank accounts and you invest $1000
in the first and $1600 in the second at the same time. The
first account pays 5% annual interest and the second 4%
annual interest, both compounded continuously. How long
will it take for the balances in the accounts to be equal?
• Hint: Use the formula A  Pert .
• Answer: 47 years
An unknown interest rate
• Suppose you make two separate deposits of $1000 each into
a savings account, one deposit per year, beginning today.
What annual interest rate r compounded continuously gives
a balance of $3750 in the account two years from today?
• Hint: Use the formula A  Pert twice and add.
• Answer: 40.55%
Five common types of mathematical models involving
exponential and logarithmic functions
1.
Exponential growth model: y = aebt, b > 0,
where b is the continuous growth rate as a decimal per
unit time (it may be expressed as a percent).
2.
Exponential decay model: y = ae–bt, b > 0,
where b is the continuous decay rate as a decimal per
unit time (it may be expressed as a percent).
( x b) 2 / c
3.
Gaussian (normal distribution) model: y  a  e
4.
a
Logistic growth model: y 
1  b  e  rt
5.
Logarithmic models: y  a  b  ln x, y  a  b  log x
Basic shapes of graphs for first three of the five models
y  3e2x
y  3e2x
Exp.Growth
Exp. Decay
ye
x2
Gaussian
Basic shapes of graphs of the last two of the five models
y
3
1  e 5 x
Logistic Growth
y  1  log x
Logarithmic model
Population growth--exponential growth model y = a∙ebt, b>0
• A population of fruit flies is experiencing exponential growth.
After 2 days there are y = 100 flies, and after 4 days there are
y = 300 flies. How many flies will there be after 5 days?
• We have 100 = a∙e2b and we solve for a  100 e2b .We substitute
this value for a in 300 = a∙e4b, obtaining 300 100e4b e2b
300 e 4b
3
 2b  e 4b  2b  e 2b
100 e
ln 3  2b  b  12 ln 3  0.5493
100  a  e 2(0.5493)  a  33.33
• We have shown that y = 33.33e0.5493t. After 5 days, there will be
33.33e0.5493(5) flies. That is, about 520 flies.
Radioactivity--exponential decay model y = a∙e–bt, b>0
• Carbon 14 dating assumes that the carbon dioxide on Earth today
has the same radioactive content as it did centuries ago. If this is
true, then the amount of 14C absorbed by a tree that grew centuries
ago should be the same as the amount of 14C absorbed by a tree
growing today. A piece of ancient charcoal contains only 15% as
much radioactive carbon as a piece of modern charcoal. How long
ago was the tree burned to make the ancient charcoal, assuming that
the half-life of 14C is 5726 years?
• First, we determine the continuous decay rate b. Continue solution
next slide.
 b(5726)
1

e
2
ln(12 )  b(5726)
ln(12 )
b
 0.0001211
 5726
More on radioactivity
• Given that 0.15 = e–bt and b = 0.0001211, solve for t.
0.15  e 0.0001211 t
ln 0.15  0.0001211t
ln 0.15
 15666years
t
 0.0001211
• That is, the piece of ancient charcoal was created about
15666 years ago.
SAT scores--Gaussian (normal distribution) model y  a  e
( x b) 2 / c
• In 2011, the SAT mathematics scores roughly followed
the normal distribution given by
y  0.0034 e
(x 514 )2 / 27,378
, 200 x  800.
Shaded area indicates that
half of the students
scored 514 or less on
their math SAT.
Spread of a virus--logistic growth model y 
a
1  b  e  rt
• On a college campus of 5000 students, one student returns
from vacation with a contagious and long-lasting flu virus.
The spread of the virus is modeled by
5000
y
, t  0.
 0 .8 t
1  4999  e
where y is the total number of students infected after t days.
• After how many days will 40% of the students be infected?
5000
2000 (0.4)(5000
)
1  4999 e 0.8 t
1  4999 e 0.8 t  2.5
1.5
 0.8t  ln
 t  10.14days
4999
Magnitude of earthquakes--Logarithmic model y  a  b  log x
• On the Richter scale, the magnitude R of an earthquake of
intensity I is given by
I
R  log ,
I0
where I0 = 1 is the minimum intensity used for comparison.
Intensity is the amplitude of waves measured by a seismograph.
• Compare the intensities of 2 earthquakes: (a) R = 4.0, (b) R = 6.3
(a) 4.0  log I  104.0  10,000  I
(b) 6.3  log I  106.3  2,000,000  I
• Therefore, the intensity of the earthquake in (b) was about 200
times as great as that of the earthquake in (a).
Problem from a recent Final Exam
• Suppose a Gross Unknown Material (GUM) is radioactive
and has a half-life of 20 days. What is the continuous decay
rate of GUM expressed as a percentage? Round your
answer to two decimal places.
• We will use the model y = ae–bt, b > 0, where b is the
continuous decay rate expressed as a decimal.
• We solve a/2 = ae–b(20).
1
2
 e20b  ln 12  20b  b  0.0347
• Therefore, the continuous decay rate is 3.47%.
Another problem from a recent Final Exam
• The number of bacteria growing in an incubation culture
increases with time according to the formula N(t) = 5200(5)t,
where t is time measured in days. After how many days will
the number of bacteria in the culture be 650,000?
• This can be formulated as N(t) = 5200(eln(5)t), so it is an
exponential growth problem.
• We solve 650,000 = 5200(5)t => (5)t =125. Now take logs,
t  log 5  log125 t  log125/log5  3.
• Therefore, it takes 3 days for the number of bacteria to be
650,000.