chapter 4 section 4.6 exponential and logarithmic equations
TRANSCRIPT
Chapter 4
Section 4.6
Exponential and Logarithmic Equations
Chapter 4
Section 4.4
Exponential and Logarithmic Equations
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Solving Equations: Examples
1. World population
1950 1960 1970 1980 1990 2000 2010 2020 2030 2040 2050
15
14
13
12
11
10
9
8
7
6
5
4
3
2
t
P(t)
1950 2.5098
1960 3.0000
1970 3.5859
1980 4.2862
1990 5.1233
2000 6.1239
2010 7.3199
2020 8.7495
2030 10.458
2040 12.500
2050 14.942
t P( t)
P1
P2
P3
(x109)
P(t) = 3(1.018)t–1960
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Solving Equations: Examples
1. World population P(t) (billions) in year t is given by
P(t) = 3(1.018)t–1960
What was the population in 1950 ? In 2000 ?
P(1950) = 3(1.018)1950–1960
P(2000) = 3(1.018)2000–1960
What is the growth factor ?
What is the average annual increase from 1950 to 2000?
From 2000 to 2050 ?
= 3(1.018)–10 = 2.50982 billion
= 3(1.018)40 = 6.12396 billion
1.018
Pt
=6.12396 – 2.50982
2000 – 1950 =
3.61414
50 = 72,282,704 people/yr
P(2050) – P(2000)
2050 – 2000 = 176,369,423 people/yr
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Solving Equations: Examples
1. World population (continued) What is the per cent change in population from 1950 to 2000 ?
What is the per cent change in population from 2000 to 2050 ?
Exponential Fact:
On equal intervals exponential growth functions always grow at a rate proportional to the initial value on the interval
NOTE: The same is true for exponential decay functions
P(2000) – P(1950)
P(1950)(100)
6.1239 – 2.5098
2.5098(100)= = 143.999 %
P(2050) – P(2000)
P(2000)(100)
14.9424 – 6.1239
6.1239(100)= = 143.999 %
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Solving Equations: Examples
1. World population (continued)
1950 1960 1970 1980 1990 2000 2010 2020 2030 2040 2050
15
14
13
12
11
10
9
8
7
6
5
4
3
2
t
P(t)
1950 2.5098
1960 3.0000
1970 3.5859
1980 4.2862
1990 5.1233
2000 6.1239
2010 7.3199
2020 8.7495
2030 10.458
2040 12.500
2050 14.942
t P( t)
P = 3.6141
t = 50
P1
P2
P3
P
P1 (100) = 143.99 %
t = 50
P = 8.8185
P
P2 (100) = 143.99 %
(x109)
Pt
=3.6141
50billion/yr
Average annual rate of change
Average annual rate of change
Pt
=8.8184
50= 176,369,423 people/yr= 72,282,704 people/yr
billion/yr
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Solving Equations: Examples
1. World population (continued)
When will the population reach 20 billion ?
World population P(t) in billions in year t is given by
P(t) = 3(1.018)t–1960
20 = 3(1.018)t–1960 Recall that logb xt = t logb x
log 20 = log (3 (1.018)t–1960 )
= log 3 + (t – 1960)log (1.018)
t = 1960 +log 20 – log 3
log 1.018
= 1960 + 106.34
≈ 2066
Question: How long till P(t) is 40 billion ?What is the doubling time for P(t) ?
… about 39 years… about 39 years
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Solving Equations: Examples
2. Solve e2x = e5x–3
ln (e2x) = ln (e5x–3)
2x = 5x – 3
3 = 3x
x = 1
3. Solve 2x = –4
Question:
What are the domain and range of f(x) = 2x ?
Domain = R
Range = (0, )
so, 2x > 0 for all x
Clearly there is no solution
Solution set: { 1 }
Solution set: { }
... OR ... use 1-1 property
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Solving Equations: Examples
4. Solve
Since the exponential function is 1-1 then
x2 = 4x – 3
x2 – 4x + 3 = 0
(x – 1)(x – 3) = 0
x – 1 = 0 OR x – 3 = 0
x = 1 OR x = 3
NOTE: Could have applied inverse function
x2 = 4x – 3
Solution set: { 1, 3 }
x2 7 74x–3 =
Zero product rule
x2 7 74x–3 =log7 log7
Question: Could we use log or ln instead of log7 ?
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Solving Equations: Examples
5. Solve 5 ln x = 10
ln x = 2
By definition 2 is the exponent of the base that yields x
Thus x = e2 ≈ (2.71828)2
≈ 7.38906
6. Solve log3 (1 – x) = 1
By definition 31 = 1 – x
Thus x = 1 – 3 = –2
Solution set: { ~ 7.38906 }
Solution set: { –2 }
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Solving Equations: Examples
7. Solve ln (x2 – 4) – ln (x + 2) = ln (3 – x)
Using the quotient rule
Since the logarithm function is 1-1
x2 – 4 = (x + 2)(3 – x)
2x2 – x – 10 = 0
2x – 5 = 0 OR x + 2 = 0
x = 5/2 OR x = –2
x2 – 4 x + 2
ln( ) = ln (3 – x)
x2 – 4 x + 2
= 3 – x
= 3x – x2 + 6 – 2x
= (2x – 5)(x + 2)
Question: What are the domain and range of the ln function ?
Solution set: { 5/2 } Why not –2 ?
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Think about it !