Chapter 5
Exponents and Logarithmic Functions
1
Overview
• Exponential functions
• y = 𝑒𝑥
• Logarithmic Functions
• Properties of Logarithms
• Inequalities and equations with exponents and logs
• Compound interest
• Exponential Growth and Decay
2
Example of Exponential
• Suppose you were paid each day for attending class.
Furthermore, suppose you were paid 2 cents on day 1, 4
cents on day 2, 8 cents on day 3, etc.
• How much will you earn on the 30th day?
y = 2x = 230  109
210  103 as an approximation
3
More Estimating
• 210 103
• 230 = 109
• 240 = 1012
4
Definitions
• Let b be an arbitrary positive number
• For every x, bx is a unique real number
• We can always estimate bx by br, where r is rational
• If bx = by and b ≠ 1 then x = y
5
Manipulation of Exponents
• (3 2 )
2
= 32 = 9
• Solve 4x = (2)2x= 8
8 = 23 so
2x = 3, x = 3/2
6
Exponential Function
• Define the exponential function:
y = bx, b > 0, b ≠ 1
• Graph of y = 2x
• Note, asymptote at y = 0, cannot be negative
7
Related Graphs
• y = 2x
• y = 2-x
• y = -2x
• y = 2-x -2
8
Solving an Equation
• X3 2x – 3(2x) = 0
9
Solution
• X3 2x – 3(2x) = 0
• Factor out 2x
• 2x(x3 – 3) = 0
• Since the exponential is >0, we have x =
3
3
10
Other Bases
• y = bx, b > 0, b ≠ 1
b>1
0<b<1
11
Horizontal Shift
• y = 2x
• y = 2x-3
12
Chapter 5.2
The Exponential Function, y = 𝑒 𝑥
13
Why e?
• First coined by Leonard Euler, ~ 1728
• e = 2.71828…, like , it is irrational
• Has many real-world applications:
• e=
∞ 1
0 𝑛!
• Probability: if a gambler plays a slot machine that pays out
with a probability of one in n and plays it n times. Then, for
large n the probability that the gambler will lose every bet
is (approximately) 1/e.
• Also used in normal distributions
14
y = ex
• Similar to y = 2x
• The number is different, but graphing, etc., is similar
15
Chapter 5.3
Logarithmic Functions
16
History
• Logarithms were introduced by John Napier in the early 17th
century as a means to simplify calculations.
• Johannes Kepler coined the notation “log”
• They were rapidly adopted by navigators, scientists,
engineers, and others to perform computations more easily
• You have missed a fair bit of “number sense” by never having
to use a slide rule.
17
Applications
• Beyond calculating, logarithmic scales have worked their way
into numerous fields
• The decibel, a logarithmic quantity, measures signal power
log-ratios, used in sound, earthquakes, and other measures
• In chemistry, pH is a logarithmic measure for acidity
• Measurements of complexity of algorithms and fractals are
given in logarithmic scales
• Logarithms describe musical intervals
• etc., etc.
18
Logarithms as Inverse Functions
• The logarithm is the inverse of the exponential
• logbx = y is the inverse of x = by
• When trying to evaluate a logarithm, can walk yourself
through the following argument:
Want to find log5125. Ask: 5 raised to what power is 125?
Clearly 5 to the 3rd power is 125, so log5125 = 3
19
Examples
Evaluate
• Log25(1/625)
• Log16(1/64)
• Log10(10)
• Log2(8 2)
• Log4(1)
20
Solution
• Log25(1/625) = -2
• Log16(1/64) = -3/2
• Log10(10) = 1
• Log2(8 2) = 3.5
• Log4(1) = 0
21
Logarithm Properties
logbx, b>1
• Domain: all reals > 0
• Range: all reals
• No y intercept, x intercept x = 1
• Asymptote: y axis
• Increasing for all x
• Note: log a/log b ≠ a/b
log a / a
22
Graph of a Logarithm
• log10x
23
Log10(x) and 10x
24
Find the Domain
• y = ln(2 – x – x2)
• y = log10
2𝑥+3
𝑥−5
25
Solution
• y = ln(2 – x – x2); need 2 – x – x2 > 0 or x2 + x- 2<0
x2 + x- 2 = (x-1)(x+2); this is <0 if only one term <0.
That will happen for -2<x<1
• y = log10
2𝑥+3
;
𝑥−5
Again, we need to have the quotient >0
numerator >0 if x>-3/2, denominator >0 if x>5
The domain is x<-3/2 or x>5
26
Example
• Solve 10 x/2 = 16
27
Solution
• Solve 10 x/2 = 16
• Take log10 of both sides
x/2 = log10 16
x = 2log10 16, = 2.408
28
Natural Logarithm
• Uses base e
• Sometimes written ln
• ln (e 2) = 2
29
Example
Simplify:
• ln e
• ln e -2
• (ln e) -2
30
Solution
Simplify:
• ln e = 1
• ln e -2 = -2
• (ln e) -2 = (1) -2 = 1
31
Other Notation
• In scientific applications, if the base is not given, it is assumed
to be base 10
• This is sometimes referred to as the “common log”
• Base 10 and e are the most commonly used bases for
most applications
32
Richter Scale
• Northridge, 1994
Magnitude 6.8
• Gugerat, 2001
Magnitude 7.9
• Magnitude M = log10(A/A0), where A is the amplitude and A0 is
a reference
• What is the relative magnitude, the ratio of the amplitude of
these two quakes?
33
Solution
• M = log10(A/A0) becomes A = A010M
• Gugerat/Northridge =
𝐴0 107.9
=
𝐴0106.8
10 1.1
• So the Gugerati earthquake was just over 10 times the
amplitude of the Northridge quake
• This is a “logarithmic” scale – one point is a power of 10
34
Reminder
• The logarithm is the inverse of the exponential
• logbx = y is the inverse of x = by
• When trying to evaluate a logarithm, can walk yourself through
the following argument:
logbx = y means b raised to what power is y
Want to find log5125? Ask: 5 raised to what power is 125?
Clearly 5 to the 3rd power is 125, so log5125 = 3
35
Chapter 5.4
Properties of Logarithms
36
Properties (and why we like logs!)
For p, q positive:
• logb b = 1
• log 1 = 0
• log pq = log p + log q
• log p/q = log p – log q
• log pr = r log p
• logb b x = x
37
Examples
Simplify – assume any base, unless stated, and assume x is in
the domain of the function:
• ½ log x – log (1+x2)
• log(x2 – 9) + 2 log
1
𝑥+3
+ 4 log x
• log 3𝑥
• log
• log
3
2𝑥
3𝑥 2 +1
𝑥 2 2𝑥 −1
2𝑥+1 3 2
38
Solutions
• ½ log x – log
•
log(x2
(1+x2)
– 9) + 2 log
log
= log 𝑥
1
𝑥+3
1 2
–
log(1+x2)
+ 4 log x = log
= log
(𝑥 2 −9)(𝑥 4 )
(𝑥+3)2
𝑥1 2
1+𝑥 2
=
(𝑥−3)𝑥 4
(𝑥+3)
• log 3𝑥 = ½(log 3 + log x)
• log
• log
3
2𝑥
3𝑥 2 +1
1
=
3
𝑥 2 2𝑥 −1
2𝑥+1 3 2
[log 2 + log x – log (3x2 + 1) ]
= 2 log x + ½ log(2x – 1) – 3
2
log(2x + 1)
39
The Slide Rule
• Developed in 17th century, based on the work in logarithms
• Useful for multiplying and dividing, but also for use in trig and
with logs; does not add or subtract
• Relies heavily on the fact that log ab = log a + log b
• Usually precise to two significant figures; user estimates the
third digit
• User must also estimate the magnitude of the result
40
The Slide Rule
• The Slide Rule's Last Century … the 20th Our last century
could not have been built without the slide rule, yet you don’t
see it
• Slide rule researchers have estimated that possibly 40 million
slide rules were produced in the world in the 20th century
alone.
– Among these are many types of specialty slide rules
developed and made for specific applications
– chemistry, surveying, electricity and electronics, artillery
ranging, hydraulics
41
The Slide Rule got us to the moon
• NASA engineers used slide rules to build the rockets and plan
the mission that landed Apollo 11 on the moon.
• It's said that Buzz Aldrin needed his pocket slide rule for lastminute calculations before landing.
• But, history marches on: In 1972 Hewlett-Packard came out
with the first handheld electronic calculator. Practically
overnight, the slide rule had become obsolete.
• With a calculator is you don't have to think about the math –
but that’s also a bad thing. When you use a slide rule you
have to estimate – be engaged in your calculations
42
Solving an Equation
Find the intercept:
• y = 2x - 3
43
Solution
find the intercept:
• y = 2x – 3
• 0 = 2x – 3
• 2x = 3
• x log 2 = log 3
• x = log 3 / log 2
NOTE: log 3 / log 2 is not log 3/2 nor log 3 – log 2
44
Examples
Solve, expressing the solutions in natural logs
• 3 𝑒 1+𝑡 = 2
• 53𝑥−1 = 27
• 102𝑥+3 = 280
45
Solution
• 3 𝑒 1+𝑡 = 2 ; 𝑒 1+𝑡 = 2/3, Take log of both sides, 1+t = ln 2/3
t = ln 2/3 – 1 = ln2 – ln3 - 1
• 53𝑥−1 = 27 ; ln 53𝑥−1 = (3x-1) ln 5 = ln 27 = 3 ln3; 3x – 1 =
3ln 3 / ln 5, 3x = 3ln3 /ln 5 + 1, x = ln3/(ln5) + 1/3
• 102𝑥+3 = 280 ; (2x+3)(ln 10) = ln 280; 2x + 3 = ln 280/ln 10,
x = ln 280/2ln10 – 3/2
46
Changing Bases
• Let z = log25
• We know 2z = 5
• If we take the log base 10 of each side, we have
log102z=log105
• But log102z = z log102
• Hence z =
𝑙𝑜𝑔10 5
𝑙𝑜𝑔10 2
• So, log25=
𝑙𝑜𝑔10 5
𝑙𝑜𝑔10 2
47
General Rule
𝑙𝑜𝑔𝑏 𝑥
• logax =
𝑙𝑜𝑔𝑏 𝑎
• To change bases, need to divide by the log of the old
base (a), expressed in the new base (b)
48
Example
• Simplify (log2 10) (log10 2)
49
Solution
• Simplify (log2 10) (log10 2)
• We want to go from base 2 to base 10 so need to divide by
log102
𝑙𝑜𝑔10 10
• log2 10 =
𝑙𝑜𝑔10 2
=
1
𝑙𝑜𝑔10 2
• Putting this in our original equation, we have
(log2 10) (log10 2) = 1
50
Examples
Express the following in Base 10
• log5 10
• ln 10
• log2 b
51
Solutions
Express the following in Base 10
• log5 10 = log10 10/log10 5 = 1/log10 5
• ln 10 = log10 10 / log10 e = 1/ log10 e
• log2 b = log10 b / log10 2
52
Things to watch for!
• log(x+y) ≠ log x + log y = log (xy)
•
log 𝑥
log 𝑦
≠ log x – log y = log (x/y)
• (log x)3 ≠ 3 log x = log x3
• log (x/2) ≠ (log x)/2 = log (x½)
53
Are the following correct?
• log a + log b – ½ log c = log(ab/ 𝑐)
• ln 𝑒 = ½
• ln x3 = ln 3x
• ln x3 = 3 ln x
• ln 2x3 = 3 ln 2x
• logac = b means ab = c
54
Solution
• log a + log b – ½ log c = log(ab/ 𝐶) T
• ln 𝑒 = ½ T
• ln x3 = ln 3x F
• ln x3 = 3 ln x T
• ln 2x3 = 3 ln 2x F
• logac = b means ab = c T
55
Chapter 5.5
Equations and Inequalities
with Logs and Exponents
56
Examples
• (ln x)2 = 2lnx
• ln(x2) = 2lnx
• Are the values x = 1 or x = 2 roots of either equation?
57
Solution
• (ln x)2 = 2lnx
• ln(x2) = 2lnx
• Are the values x = 1 or x = 2 roots of either equation?
• In the first:
ln 1 = 0, and 02 = 2(0), true
(ln2) 2 = 2ln 2; dividing by ln2, we have ln2 = 2, incorrect
• In the second:
ln (x2) = 2 lnx for all x, so holds for both values
58
Example; variable in exponent
• 5x = 43x+5, find x
59
Solution
• 5x = 43x+5, find x
• Take logs of both sides
ln(5x) = ln(43x+5)
• This gives:
x ln5 = (3x+5) ln 4
x ln 5 – 3x ln4 = 5 ln4
x = (5 ln4) / (ln5 – 3 ln4)
60
Example
Solve for x
• ln(ln x) = 2
• e ln x = -2
• e ln x = 2
61
Solution
Solve for x
• ln(ln x) = 2 ; let z = ln x; Then the equation becomes ln z = 2;
this is the same as z = e2, and the equation in just ln x = e2, or
x = e e2
• e ln x = -2
et is always > 0, so has not solution
• e ln x = 2
e ln x = x for all x, so x = 2
62
Some Review
• If f (x) = 𝑒 𝑥+1 find its inverse and specify the domain
• What is the domain of ln (x+2)
63
Solution
• If f (x) = 𝑒 𝑥+1 find its inverse and specify the domain
y = ex+1
ln y = x+1
x = -1+ln y, or y = -1 + ln x, domain is x > 0
• What is the domain of ln (x+2)
Need (x+2) / 0, x > -2
64
More Review
• log12 2 + log12 18 + log12 4 =
• log1/2 (1/16) – log5 0.02 =
• 6 log10 [B/(A 𝐶)] =
• log3 54 – log3 2
• ln 100 / ln 10
65
Solution
• log12 2 + log12 18 + log12 4 =
Log12(2(18)(4))= log12(144) = 2
• log1/2 (1/16) – log5 0.02 =
4 – log5 (2/100) = 4-log5 (1/50) = 4 - log5 (1/2)(1/25)= 4 + 2
+log5(2)
• 6 log10 [B/(A 𝐶)] =
6[ log B – log(A 𝐶) ] = 6[log B – log A – ½ log C]
• log3 54 – log3 2
Log3(9(6)) – log32= log327(2) – log32 = log3 2 + log327 – log3 2 =
3
• ln 100 / ln 10 = ln 102/ln10 = 2 ln 10/ln 10 = 2
66
Yet More Review
• ln
•
𝑥 − 3 (𝑥 − 4) =
𝑥 2 2𝑥+1
ln
2𝑥+1
Find the domain of
• y=
• y=
2+𝑙𝑛𝑥
2 −𝑙𝑛𝑥
𝑙𝑜𝑔10 𝑥
Find the range of
• y=
𝑒 𝑥 +1
𝑒 𝑥 −1
67
Solution
• ln
𝑥 − 3 (𝑥 − 4) =1/2 ln (x-3)(x-4) , can break into a sum if
wish
•
𝑥 2 2𝑥+1
ln
=
2𝑥+1
2 lnx + ln (2x+1)1/2 = 2 lnx + ½ ln(2x+1)
Find the domain of
• y=
• y=
2+𝑙𝑛𝑥
2 −𝑙𝑛𝑥
need x > 0 and ln x 2
𝑙𝑜𝑔10 𝑥 = ½ log10x, x > 0
68
Find the Range
• y=
𝑒 𝑥 +1
𝑒 𝑥 −1
• (ex – 1)y = ex + 1
• ex (y-1) = 1+y
• ex = (1+y)/(y-1)
• x = ln[(1+y)/(y-1)],
• Domain is y ≠1, but also need (1+y)/(y-1)>0
numerator >0 for y>-1, denominator for y>1, or positive y>1
numerator <0 for y<-1, denominator for y<1, or positive y< -1
• Domain is |y|> 1 so this is the range of the inverse function
69
Graph
y=
𝑒 𝑥 +1
𝑒 𝑥 −1
6
4
2
0
-4
-3
-2
-1
-2
0
1
2
3
4
-4
-6
70
More on Equation Solving
71
Example
Solve for x, base is 10
• log (x2 – 4) = 5
• log(x2 – 4x) = 5
72
Solution
Solve for x, base is 10
• log (x2 – 4) = 5
x2 – 4 = 105, x = 105 + 4, take sqrt of both sides
• log(x2 – 4x) = 5
x2 – 4x = 105; this is a quadratic -- solve as such
73
Example
Solve:
• 7-4x = 21-3x
• log2 (2x2 – 4) = 5
• 3(22x) – 11(2x) – 4 = 0
• ln(ln(ln(x))) = 1
74
Solutions
Solve:
• 7-4x = 21-3x
log(7-4x) = log(21-3x), -4x log7 = (1-3x)log 2
x(-4 log7 + 3log 2) = log2; x = -log2/(4log7 + 3 log2)
• log2 (2x2 – 4) = 5
(2x2 – 4) =25; x=± 24 + 2 = ± 18 = ±3 2
• 3(22x) – 11(2x) – 4 = 0
let y = 2x,3y – 11y – 4 = 0, (3y+1)(y- 4) = 0; y = -1/3, 4
2x=4, x = 2, 2x= -1/3, no solution; x = 2
• ln(ln(ln(x))) = 1; ln(y) = 1 then y = e, so ln(ln(x)) = e; raise both
sides to power e, ln(x) = ee; x = eee
75
Examples
Solve for x:
• e3x = 102x(21-x)
• ln x + ln (x+1) = ln 12
• ln (3x2) = 2 ln( 3 x)
• log(x+1) = 2 log(x-1)
76
Solution
Solve for x:
• e3x = 102x(21-x)
3x = ln(102x(21-x))= 2x ln10 + (1-x)ln2
x(3 – 2ln10 + ln2) = ln2; x = ln2/(3 – 2 ln10 + ln2)
• ln x + ln (x+1) = ln 12
ln(x(x+1)) = ln12; x(x+1) = 12, x = 3
•
ln (3x2) = 2 ln( 3 x)
ln (3x2) =ln(3x2), all x > 0
• log(x+1) = 2 log(x-1)
log(x+1) = log(x-1)2; x+1 = (x-1)2; x2 – 3x = 0; x = 3, x = 0, but
can’t have x = 0 (log(x-1) not allowed)
77
Inequalities
• Solve just like equalities:
2(1+0.4x)<5
78
Solution
• 2(1+0.4x)<5
(1+0.4x)<5/2
0.4x< 3/2
ln(0.4x)= x ln (0.4)< ln(3/2)
We need to divide by ln(0.4), but ln(0.4) is negative. How do
we know that? ln(1) = 0, and, since the function ln(x) is
always increasing, ln(a) for a<1 is < 0
x > ln(3/2)/ln(0.4)
79
Example
Solve:
• ln(2-3x) ≤ 1
• e2-3x ≤ 1
80
Solution
Solve:
• ln(2-3x) ≤ 1
2-3x ≤ e
3x ≥ 2-e
x ≥ (2-e)/3
Verify domain: need x < 2/3, so (2-e)/3 ≤ x < 2/3
• e2-3x ≤ 1
2-3x ≤ log(1) = 0
3x ≥ 2, x ≥ 2/3
81
Solve
• Let log be base 10, solve for x
• log x + log (x-2) ≤ log 24
82
Solution
• log x + log (x-2) ≤ log 24
log(x)(x-2) ≤ log 24
x(x-2) = 24, x2 – 2x – 24 = 0
x = 6, x = - 4
First check domain: we have to have x >2, so can’t use x = -4
Need x(x-2) ≤ 24, 6 is the key point. Try x = 5: 5(5-2) =
15 ≤ 24; try x = 7: 7(7-2) = 35 . 24,
Solution is 2<x ≤ 6
83
Examples
• 2/3(1-e-x)<-3
• ln
3𝑥 −2
4 𝑥+1
> ln 4
• e1/(x-1) > 1
84
Solutions
• 2/3(1-e-x)<-3
(1-e-x) < -9/2; e-x> 11/2; so ex> 2/11. Take log of both sides
x > ln (2/11)
• ln
3𝑥 −2
4 𝑥+1
> ln 4
3𝑥 −2
>
4 𝑥+1
4; 3x-2 > 16x + 1; 13x < -3, x < -3/13; check
domain: need (3x-2)/(4x+1)>0; -6/13 < x < -1/4
• e1/(x-1) > 1 take ln of both sides
1/x-1 > 0; Need x > 1
85
Watch for Extraneous Roots
• Solve log3x + log3(x+2) = 1
• log3 (x2 + 2x) = 1
• (x2 + 2x) = 3
• (x+3)(x-1) = 0, x = -3, x = 1
• Check results:
for x = -3, we have log3 (-3), which is not defined
86
Example
• 3(ln x)2 – ln (x2) – 8 = 0
87
Solution
• 3(ln x)2 – ln (x2) – 8 = 0
• Rewrite 3 (lnx)2 – 2lnx – 8 = 0
• Substitute and treat as a quadratic
• y = lnx
• 3y2 – 2y – 8 = 0
• (3y + 4)(y-2), y = -4/3, y = 2
• y = lnx, so x = ey, x = e -4/3, x = e2
• Check: 3(4) – 4 – 8 = 0, e2 works
3(ln e -4/3)2 – ln e -8/3 -8 = 3(4/3)2 – 8/3 – 8 = 12/3 – 8/3 – 8 ≠0
so -4/3 doesn’t work.
88
Example
• Log6x =
1
1
1
+
𝑙𝑜𝑔2 𝑥 𝑙𝑜𝑔3 𝑥
89
• Log6x =
1
1
1
+
𝑙𝑜𝑔2 𝑥 𝑙𝑜𝑔3 𝑥
• Rewrite logax = logbx/logba
• Denominator becomes (log62 + log63)/log6 x = log6x / log66
= log6x, since logaa=1
• The equations holds for x>0, x≠1
(if x = 1, we divide by 0, since log 1 = 0)
90
Section 5.6
Compound Interest
91
Compound Interest
• Suppose you invested $100 and the interest was 5%
• The first time you got an interest payment, you have $105
• The second time, you have $105(1.05) =$110.25, already
more than a 10% increase.
92
Terminology
• Principal: the amount you invest, P
• Compounded annually: every year, you earn 10% on your
original deposit.
• In general, if the interest rate is r,
– The first year you have P(1+r)
– The second year you have P(1+r)(1+r)
– The third year you have P(1+r)(1+r)(1+r)
– At the end of t years, you have P(1+r)t
93
Example
• Suppose you invest $5000 at 3.5% interest
compounded annually. How many years will it take for your
money to double?
94
Solution
• Suppose you invest $5000 at 3.5% interest
compounded annually. How many years will it take for your
money to double?
5000 (1+0.035)t = 10,000
(1.035)t = 2
log (1.035)t = log 2
t log (1.035) = log 2
t = log2/ log (1.035) = 20.7 years
95
However
• Usually interested is compounded more than once a year
• If interest is compounded n times a year, the periodic interest
rate is r/n
• The amount earned A is
A = P(1 + r/n)nt, where nt is the number of periods, n per year
for t years
96
Example
• Suppose you invest $1000 at 10% per year.
• How much more do you earn if interest is compounded
quarterly than if it is compounded annually?
• 1000(1+0.1/1)(1)(1) = $1100
• 1000(1+0.1/4)(1)(4) = $1103.81
• We call the effective rate 10.381% vs. the nominal rate of 10%
97
Solution
• 1000(1+0.1/1)(1)(1) = $1100
• 1000(1+0.1/4)(1)(4) = $1103.81
98
Interesting Numbers
Results of Compounding Interest More Frequently
Number of
Compoundings, n
Amount (1 + 1/n)n
1 (annually)
2
2 (semiannually)
2.25
4 (quarterly)
2.44
12 (monthly)
2.61
365 (daily)
2.7146
8760 (hourly)
2.7181
525,600 (each minute)
2.71827
99
Continuous Compounding
• A = Per t
Here r is the annual rate, t is the number of years
• What is the doubling time?
•
100
Solution
• A = Per t
• 2P = Per t
• 2 = er t
• ln2 = rt
• t = (ln2)/r
• Note that, as expected, the time required to double is
independent of the amount invested
101
Example
• Assume 4% interest per year
• t = (ln2)/r
• t = (ln 2)/0.04 = 17.3 years
• ln2  0.7, thus if we have the rate, it is easy to determine the
doubling time, and vice versa
If r = 10%, doubling takes 7 years
102
Section 5.7
Exponential Growth and Decay
103
Meaning
• “Exponential” means governed by functions of the form
N = N0 ekt
N is the number
N0 is the initial number
y is time
k is the growth (or decay) constant
Many natural phenomena follow such a pattern: population
growth, radioactive decay,
104
More on Constants
• For N = N0 ekt
• If k >0, grows
• If k < 0, decays
105
Example
• Half-Life: The half-life of a radioactive substance is the time
required for one half the substance to decay
• Iodine-131 has a half-life of 8 days. What is its decay
constant, k?
N = N0 ekt
106
Solution
• Half life is 8 days
• ½ = e 8k
• 8k = ln (1/2)
• k = (-ln2)/8 = -0.08664
107
A decay problem
• In 2000, 50 grams of radium were stored. The half-life of
radium is 1,620 years. How many grams of radium remains
after 4860 years?
108
Amount = 50(1-0.5)y
Half Life
1
1620 yr
2
3240 yr
3
4860 yr
25 gm
12.5 gm
6.25 gm
109
Carbon 14 Dating
• Prehistoric paintings were discovered that had 20% of the
original carbon-14 in them.
• Carbon-14 decays with a constant of 0.000121.
• How old are the paintings?
110
Solution
• N = N0 e-kt
• N = 0.20 (20%) of N0
• 0.2 = e-0.000421t
• Ln 0.2 = -0.000121t
• -1.61 = -0.000121t
• t = 13,301 yrs
111
Half-Life and Decay Constant
• N/N0 = ½ = ekt where k is the decay constant and
t is the half-life
• Therefore, k = ln(1/2) / half-life = -0.693 / half-life
112
Chernobyl
• This accident released cesium-137, iodine-131 and
strontium-90
• In Lapland, pastures were contaminated by cesium-137,
which has a half-life of 33 years.
• If the levels were found to be 10 times normal, how long
would it take for the levels to return to normal?
113
Solution
• Half-life is 33 years
• 10 times normal
• Decay constant is -0.693 / half-life = -0.021
• 0.1 = e(-0.021t)
• -0.021t = ln 0.1
• t = 110 yrs
114
Fable of the Chess Board and Grains of Wheat
There is a well-known fable about a man from India who
invented the game of chess, as a gift for his king. The king was
so pleased with the game that he offered to grant the man any
request within reason. The man asked for one grain of wheat to
be placed on the first square of the chess board, two grains to be
placed on the second square, four on the third, eight on the
fourth, etc., doubling the number of grains of wheat each time,
until all 64 squares on the board had been used. The king,
thinking this to be a small request, agreed.
115
Chess Board, Continued
A chess board has 64 squares.
How many grains of wheat did the king have to place
on the 64th square of the chess board?
116
• y = 2n ; The number of grains on square n; need to add 1 + 2
+ 4 + … 264
• What is 264? 1.8 x 1019
• There would be a total of 3.7x 1019 grains of wheat on
the board
117
How much wheat is this?
China produces 3.8 bushels per year. It would take the output of
6000 years to fill the chess board
118