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) = 2 x . The table below lists some values of the function.
x f(x) –4 1/16 –3 1/8 –2 1/4 –1 1/2 0 1 • Note that f is an increasing function.
1 2 2 4 3 8 4 16
Exponential functions and their graphs, continued • The domain of f(x) = 2 x 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 some values of the function.
2 x
.
x g(x) –4 16 –3 8 –2 4 –1 2 0 1 • Note that g is a decreasing function.
1 1/2 2 1/4 3 1/8 4 1/16
Exponential functions and their graphs, continued 2 x extended to include all real numbers and the resulting graph is shown below.
Note that the x-axis is again a horizontal asymptote.
The relation between Example 1 and Example 2 • In Example 1, f(x) = 2 x while in Example 2, g(x) = 2 x
.
2
1 2 ,
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
2 x
y
2
x
Comparing graphs of exponential functions • Consider f(x) = 2 x and h(x) = 4 x . 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 4 x y 2 x
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: a x = a y if and only if x = y.
• The one-to-one property can be used to solve simple exponential equations.
• Solve 8 = 2 x+1 . Rewrite as 2 3 = 2 x+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) = 3 x . 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.
asymptote of 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 3 t 0 1 2 Balance after each compounding P P P 1 P(1 r) P 2 P 1 (1 r) P(1 r)(1 r) P(1 r) 2 P 3 P 2 (1 r) P(1 r) 2 (1 r) P(1 r) 3 P t 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 n nt .
• As the number of compoundings per year increases we have P 1 r n nt Pe rt as n .
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 (4)(10) $ 1643 .
62 • (b) We have P = 1000, r = 0.05, n =12, t = 10 A 1000 1 0.05
(12)(10) 12 $ 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 1000 x 3 1000 x 2 1000 x • To find x we must solve 1000 x 3 1000 x 2 1000 x 3300 • This means we must find a zero of the polynomial Q(x) x 3 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
Q
200(0.886)
5.727
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 log a x a y if 1 and only if x a y .
The function given by f(x) log a 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 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: • log 2 64 • log 2 0.5
• log 3 1 • l
og
3
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 0.01 0.1 1 10 100 1000 log x -2 -1 0 1 2 3 • Using Maple or graphing calculator, we can plot the graph of log x:
Properties of logarithms 1. log a 1 = 0 because a 0 = 1.
2.
3.
log a a = 1 because a 1 log a a x = a.
log a x 4.
If log a x = log a y, then x = y (one-to-one property)
One way : log
4
x
2 4 x 2
log
4
16
2
x
2
16
x
4.
2nd way : 4
log 4 x 2
4
2
x
2
16
x
4.
Property 3 may be interpreted as follows: If f(x) a x , then f 1 ( x) log a x.
Typical logarithm graph • When a > 1, a typical graph of log a x is shown along with some of its properties.
Domain: (0, ) y log a x Range: (– , ) x-intercept: (1,0) x Increasing One-to-one => inverse exists y-axis is a vertical asymptote Continuous Reflection of graph of y = a x 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 $ 0 .
01 10 2 .
We note that the order of magnitude is the logarithm of the ratio of their values.
Graphs of 10 x and log x y 10 x · (0.3010, 2) (–1, 0.1) · (0,1) (1,0) · (0.1, –1) (2, 0.3010) · y log x
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 1 y log (x 1) x 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) log e 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 e 2 = 2. Otherwise, use a calculator. For example, ln 2 = .6931.
Properties of natural logarithms 1. ln 1 = 0 because e 0 = 1.
2.
ln e =1 because e 1 = e.
3.
ln e x = x and e ln x = x. (inverse properties) 4.
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 : e
ln(x 1 )
e
1
x
1
e
1
x
1
e
1
.
Property 3 may be interpreted as follows: 1 If f(x) e x , then f ( 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 positive real numbers such that a 1 and b 1.
Then can be converted to a different base as follows. a Base b Base 10 Base e log a x log log b x log b a a x log x log log a a x ln x ln a • Example. Evaluate log 4 25 using both common and natural logarithms.
log 4 25 log 25 log 4 1 .
39794 0 .
60206 2 .
3219 log 4 25 ln ln 25 4 3 .
21888 1 .
38629 2 .
3219
Properties of logarithms, continued real number. If u and v are positive real numbers, the following properties are true.
1.
Product Property : log a (uv) log a u ln(uv) log ln u ln v a v 2.
Quotient Property : log a u v log a u log a v 3.
Power Property ln : log a u u v ln u ln v n n log a u ln u n n ln u
Rewriting logarithmic expressions using properties of logs • Expand the given expression.
log x 2 x 3 1 log( x 2 1 ) log x 3 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/m 2 is compared to a standard benchmark sound, I 0 . This results in the following definition: where I 0
Noise level in decibels
10 is defined to be 10 -12 log I I 0 , watts/m 2 , roughly the lowest intensity audible to humans.
• Problem. If a sound doubles in intensity, by how many units does its decibel rating increase?
Difference in decibel ratings 10 log 2I I 0 10 log 2I I 0 log I I 0 10 log 10 log I I 0 3 .
010 dB.
• Solving exponential and logarithm equations The following strategies are available, but strategies 2 and 3 are the most important.
1.
2.
3.
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.
Example for 1. 2 x = 64 => 2 x = 2 6 => x = 6 Example for 2. 2 x = 64 => x∙log 2 = log 64 => x = log 64/log 2 = 6 alternatively, 2 x = 64 => log 2 2 x = log 2 64 => x = log 64/log 2 = 6 Example for 3. log x = –2 => 10 log 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
(1/4)
t
20, by subtractin g
1/6, by dividing log(1/4)
t
log(1/6), by taking logs t t
log(1/4)
log(1/6), using a log property log(1/6)/l og(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 t log 2 log 1.06
0 .
30103 0 .
025306 11 .
896 years.
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.
log
1.06
1 .
06
t
log
1.06
2 t
log 2 log 1.06
11 .
896 years, using change of base
• 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 ) e 1 exponentia te x(x 2) e inverse prop.
x 2 2x e 0 algebra x 1 1 e quadratic formula x 1 1 e other " solution" is negative wh ich is impossible in original equation
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 log 2 x log 2 ( x 2) log 2 ( x 6).
• Answer: x = 2 is the only answer.
Solving an exponential equation • Solve
2
( x 2 )
4
x
.
• 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 solved for t 1 2 ( 0 .
886 ) t divide by 200 log 1 2 t log 0.886
take logs, t log 0.5
/ log 0 .886
5.726
use power prop.
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 Pe rt
.
• 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?
• Answer: 40.55%
Five common types of mathematical models involving exponential and logarithmic functions 1.
Exponential growth model: y = ae bt , 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).
3. Gaussian (normal distribution) model: y a e ( x b) 2 / c 4. Logistic growth model: y 1 a b e rt 5. Logarithmic models: a b ln x, y a b log x
Basic shapes of graphs for first three of the five models y 3e 2x
Exp.
Growth
y e x 2
Gaussian
y 3e 2x
Exp
.
De cay
Basic shapes of graphs of the last two of the five models y 1 3 e 5 x
Growth
y 1 log x
c m ode l
Population growth--exponential growth model y = a∙e bt , 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∙e 2b 2b and we solve for We substitute this value for a in 300 = a∙e 4b , obtaining 300 100 e 4b e 2b 3 300 e 4b 100 e 2b ln 3 2b b 1 2 e 4b 2b e 2b ln 3 0.5493
100 a e 2(0.5493) a 33.33
• We have shown that y = 33.33e
0.5493t
. After 5 days, there will be 33.33e
0.5493(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 14 C absorbed by a tree that grew centuries ago should be the same as the amount of 14 C 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 14 C is 5726 years?
• First, we determine the continuous decay rate b. Continue solution next slide.
1 2 e b(5726) ln( 1 2 ) b(5726) b ln( 1 2 ) 5726 0 .
0001211
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 .
0001211 t t ln 0.15
0 .
0001211 15666 years • 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 1 b a 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 y 1 5000 4999 e 0 .
8 t , t 0.
where y is the total number of students infected after t days.
• After how many days will 40% of the students be infected?
2000
1
4999 (0.4)(5000
e
0 .
8 t
)
2 .
5
1
5000 4999
e
0 .
8 t
0 .
8 t
ln 1.5
4999
t
10.14
days
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 R log I I 0 , where I 0 = 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 10 4.0
10 , 000 I (b) 6.3
log I 10 6.3
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 20 b ln 1 2 20 b 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(e ln(5)t ), so it is an exponential growth problem.
• We solve 650,000 = 5200(5) t t log 5 log 125 t log => (5) t 125/log 5 =125. Now take logs, 3.
• Therefore, it takes 3 days for the number of bacteria to be 650,000.