Chapter 4 Exponential and Logarithm Functions

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Chapter 4Exponential and Logarithm Functions
4.1 Exponential Functions 4.2 The Natural
Exponential Function 4.3 Logarithm
Functions 4.4 Logarithmic Transformations 4.5
Logistic Growth
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Natural Exponential Function (Section 4.2) f(x)
ex (2.718..)x
x -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0
yex 1/e2 1/e(3/2) 1/e 1/ve e0 ve e e(3/2) e2
.1353 .2231 .3679 .6065 1 1.6487 2.7183 4.4817 7.3891
f(x) ex domain reals range positive reals y
intercept 1 increasing a gt 0
PRACTICE 15/194, 19/194
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Standard Form vs Base-e Form
f(x) Cekx base-e form
f(x) Cax standard form
growth rate is easy.
rate of change is easy.
Laws of Exponents ekx (ek)x
ax (ek)x a ek
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Standard Form vs Base-e Form
f(x) Cekx base-e form
f(x) Cax standard form
Rewrite the function f(t) 5230e0.01725t
in standard form.
e0.01725 1.0174 So f(t) 5230(1.0174)t
Rewrite the function B(t) 3000e-0.22314t
in standard form.
e-0.22314 .80000 So B(t) 3000(0.80)t
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Standard Form vs Base-e Form
f(x) Cekx base-e form
f(x) Cax standard form
Rewrite the function f(x) 1004x in base-e
form.
ek 4
Using graph and trial/errork 1.4, ek 4.0552
k 1.39, ek 4.0149 k 1.388, ek 4.0068k
1.387, ek 4.0028k 1.3865, ek 4.0008k
1.3863, ek 4.0000 So f(x) 100e1.3863t
There must be an easier way! (how?)
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Standard Form vs Base-e Form
Rewrite the function f(x) 2x in base-e form.
Rewrite the function f(x) 3x in base-e form.
Since e0.6931 2 (how?), then f(x) 2x
e0.6931x
Horizontal stretch by a factor of 1/(0.6931)
1.442695
Since e1.0986 3 (how?), then f(x) 3x
e1.0986x
Horizontal stretch by a factor of 1/(1.0986)
0.91204
Every f(x) ax can be viewed as a horizontal
stretch/compression of g(x) ex .
It is possible to live your whole life with just
the natural exponential function!!!
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How do we solve for an unknown variable appearing
as an exponent?
ek 4
How do we find input corresponding to given
output for an exponential function?
y ax
We need the inverse function of an exponential
function.
LOGarithm Functions are the inverse functions
of exponential functions.
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Section 4.3 The Logarithm Function
The inverse function for f(x) ax is given by f
-1 (x) loga(x)
loga(x) y ay x
loga(x) recognizes x as a power of a.
Practice CYU 4.9/176
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The Common Logarithm
The inverse function for f(x) 10x Is given by f
-1 (x) log10(x) log(x)
log10(x) y 10y x
log10(x) recognizes x as a power of 10.
Practice CYU 4.8/174
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The Common Logarithm (4.3)
x -2.0 -1.0 0.0 1.0 2.0
y10x 1/100 1/10 1 10 100
x 1/100 1/10 1 10 100
ylog10(x) -2.0 -1.0 0.0 1.0 2.0
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The Natural Logarithm
The inverse function for f(x) ex is given by f
-1 (x) loge(x) ln(x)
ln(x) y ey x
ln(x) recognizes x as a power of e.
PRACTICE CYU 4.10/178
PRACTICE 23/194
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The Natural Logarithm (4.3)
x -2.0 -1.0 0.0 1.0 2.0
yex 1/e2 1/e e0 e e2
.1353 .3679 1 2.7183 7.3891
x .1353 .3679 1 2.7183 7.3891
yln(x) -2.0 -1.0 0.0 1.0 2.0
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The Family of Logarithmic Functions
as x?8, loga(x)? 8 as x?0, loga(x)? -8
as x ? 8, ax ? 8 as x ? -8, ax ? 0
f(x) ax with a gt 1 domain reals range
positive reals HA y 0 y intercept 1
f(x) loga(x) with a gt 1 domain positive
reals range reals VA y 0 x intercept 1
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The Family of Logarithmic Functionsa gt 1
as x?8, loga(x)? 8 as x?0, loga(x)? -8
f(x) loga(x) with a gt 1 domain positive
reals range reals VA y 0 x intercept 1
PRACTICE 25/194, 27/194, 31/195, 33/195,
37a,d/195
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Applications of Logarithms (4.3)
f(x) Cekx base-e form
f(x) Cax standard form
ek 4 Using graph and trial/errork 1.4, ek
4.0552 k 1.39, ek 4.0149 k 1.388, ek
4.0068k 1.387, ek 4.0028k 1.3865, ek
4.0008k 1.3863, ek 4.0000 So f(x)
100e1.3863t
Rewrite the function f(x) 1004x in base-e
form.
ek 4 ln(ek) ln(4) k ln(4) 1.3863 So
f(x) 100e1.3863t
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Applications of Logarithms (4.3)
Example Rewrite the cooling coffee function in
base-e form.
H(t) 20 70(0.963)t
H(t) 20 70(e)kt
(e)kt (0.963)t
(e)k (0.963) ln(ek) ln(0.963) k ln(0.963)
-0.03770
H(t) 20 70(e)-0.03770t
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through Section 4.3/page 179 HW 15-38 on pages
194-5 TURN IN 16a (by hand graph), 20a, 25a
(by hand graphs), 28, 34, 37b,c
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WHAT ABOUT e? (4.2)
Suppose we want to determine the rate of change
of a function at an instant.
Suppose we want to determine the instantaneous
rate of change of f.
For any linear function, the average rate of
change and instantaneous rates of change are
always constant and equal to the slope.
For curvy functions, this question of rate of
change at an instant or slope at a point is a
BIG DEAL!
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WHAT ABOUT e? (4.2)
Example In the cooling coffee example, how fast
is the coffee cooling 10 minutes after being
poured, when its temperature is 68C?
H(t) 20 70(0.963)t
looks linearrate of change is constant
Ten minutes after being poured, the coffee is
cooling at a rate of 1.8102 degrees C per minute
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WHAT ABOUT e? (4.2)
Consider the instantaneous rates of change for
two simpler exponential functions.
f(x) 3x
f(x) ex
point instantaneous rate of change(quotient) instantaneous rate of change(calculation)
(-2, 1/9) .1221
(-1, 1/3) .3662
(0, 1) 1.0986
(1, 3) 3.2959
(2, 9) 9.8877
point instantaneous rate of change(quotient) instantaneous rate of change(calculation)
(-2, 0.1353 ) .1353
(-1, 0.3679) .3679
(0, 1) 1.0000
(1, 2.7183) 2.7183
(2, 7.3891) 7.3892
What do you observe?
rate of change matches output value for the
natural exponential function
the natural exponential function is the ONLY
function with this property