general logarithms and exponentials
TRANSCRIPT
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General Logarithms and Exponentials
Last day, we looked at the inverse of the logarithm function, the exponentialfunction. we have the following formulas:
ln(x)
ln(ab) = ln a+ln b, ln(a
b) = ln a−ln b
ln ax = x ln a
limx→∞
ln x =∞, limx→0
ln x = −∞
d
dxln |x | =
1
xZ1
xdx = ln |x |+ C
ex
ln ex = x and e ln(x) = x
ex+y = exey , ex−y =ex
ey, (ex)y = exy .
limx→∞
ex =∞, and limx→−∞
ex = 0
d
dxex = ex
Zexdx = ex + C
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 2: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/2.jpg)
General exponential functions
For a > 0 and x any real number, we define
ax = ex ln a, a > 0.
The function ax is called the exponential function with base a.Note that ln(ax) = x ln a is true for all real numbers x and all a > 0. (We sawthis before for x a rational number).Note: The above definition for ax does not apply if a < 0.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 3: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/3.jpg)
Laws of Exponents
We can derive the following laws of exponents directly from the definition andthe corresponding laws for the exponential function ex :
ax+y = axay ax−y =ax
ay(ax)y = axy (ab)x = axbx
I For example, we can prove the first rule in the following way:
I ax+y = e(x+y) ln a
I = ex ln a+y ln a
I = ex ln aey ln a = axay .
I The other laws follow in a similar manner.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 4: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/4.jpg)
Laws of Exponents
We can derive the following laws of exponents directly from the definition andthe corresponding laws for the exponential function ex :
ax+y = axay ax−y =ax
ay(ax)y = axy (ab)x = axbx
I For example, we can prove the first rule in the following way:
I ax+y = e(x+y) ln a
I = ex ln a+y ln a
I = ex ln aey ln a = axay .
I The other laws follow in a similar manner.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 5: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/5.jpg)
Laws of Exponents
We can derive the following laws of exponents directly from the definition andthe corresponding laws for the exponential function ex :
ax+y = axay ax−y =ax
ay(ax)y = axy (ab)x = axbx
I For example, we can prove the first rule in the following way:
I ax+y = e(x+y) ln a
I = ex ln a+y ln a
I = ex ln aey ln a = axay .
I The other laws follow in a similar manner.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 6: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/6.jpg)
Laws of Exponents
We can derive the following laws of exponents directly from the definition andthe corresponding laws for the exponential function ex :
ax+y = axay ax−y =ax
ay(ax)y = axy (ab)x = axbx
I For example, we can prove the first rule in the following way:
I ax+y = e(x+y) ln a
I = ex ln a+y ln a
I = ex ln aey ln a = axay .
I The other laws follow in a similar manner.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 7: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/7.jpg)
Laws of Exponents
We can derive the following laws of exponents directly from the definition andthe corresponding laws for the exponential function ex :
ax+y = axay ax−y =ax
ay(ax)y = axy (ab)x = axbx
I For example, we can prove the first rule in the following way:
I ax+y = e(x+y) ln a
I = ex ln a+y ln a
I = ex ln aey ln a = axay .
I The other laws follow in a similar manner.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 8: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/8.jpg)
Laws of Exponents
We can derive the following laws of exponents directly from the definition andthe corresponding laws for the exponential function ex :
ax+y = axay ax−y =ax
ay(ax)y = axy (ab)x = axbx
I For example, we can prove the first rule in the following way:
I ax+y = e(x+y) ln a
I = ex ln a+y ln a
I = ex ln aey ln a = axay .
I The other laws follow in a similar manner.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 9: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/9.jpg)
Derivatives
We can also derive the following rules of differentiation using the definition ofthe function ax , a > 0, the corresponding rules for the function ex and thechain rule.
d
dx(ax) =
d
dx(ex ln a) = ax ln a
d
dx(ag(x)) =
d
dxeg(x) ln a = g ′(x)ag(x) ln a
I Example: Find the derivative of 5x3+2x .
I Instead of memorizing the above formulas for differentiation, I can justconvert this to an exponential function of the form eh(x) using thedefinition of 5u, where u = x3 + 2x and differentiate using the techniqueswe learned in the previous lecture.
I We have, by definition, 5x3+2x = e(x3+2x) ln 5
I Therefore ddx
5x3+2x = ddx
e(x3+2x) ln 5 = e(x3+2x) ln 5 ddx
(x3 + 2x) ln 5
I = (ln 5)(3x2 + 2)e(x3+2x) ln 5 = (ln 5)(3x2 + 2)5x3+2x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 10: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/10.jpg)
Derivatives
We can also derive the following rules of differentiation using the definition ofthe function ax , a > 0, the corresponding rules for the function ex and thechain rule.
d
dx(ax) =
d
dx(ex ln a) = ax ln a
d
dx(ag(x)) =
d
dxeg(x) ln a = g ′(x)ag(x) ln a
I Example: Find the derivative of 5x3+2x .
I Instead of memorizing the above formulas for differentiation, I can justconvert this to an exponential function of the form eh(x) using thedefinition of 5u, where u = x3 + 2x and differentiate using the techniqueswe learned in the previous lecture.
I We have, by definition, 5x3+2x = e(x3+2x) ln 5
I Therefore ddx
5x3+2x = ddx
e(x3+2x) ln 5 = e(x3+2x) ln 5 ddx
(x3 + 2x) ln 5
I = (ln 5)(3x2 + 2)e(x3+2x) ln 5 = (ln 5)(3x2 + 2)5x3+2x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 11: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/11.jpg)
Derivatives
We can also derive the following rules of differentiation using the definition ofthe function ax , a > 0, the corresponding rules for the function ex and thechain rule.
d
dx(ax) =
d
dx(ex ln a) = ax ln a
d
dx(ag(x)) =
d
dxeg(x) ln a = g ′(x)ag(x) ln a
I Example: Find the derivative of 5x3+2x .
I Instead of memorizing the above formulas for differentiation, I can justconvert this to an exponential function of the form eh(x) using thedefinition of 5u, where u = x3 + 2x and differentiate using the techniqueswe learned in the previous lecture.
I We have, by definition, 5x3+2x = e(x3+2x) ln 5
I Therefore ddx
5x3+2x = ddx
e(x3+2x) ln 5 = e(x3+2x) ln 5 ddx
(x3 + 2x) ln 5
I = (ln 5)(3x2 + 2)e(x3+2x) ln 5 = (ln 5)(3x2 + 2)5x3+2x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 12: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/12.jpg)
Derivatives
We can also derive the following rules of differentiation using the definition ofthe function ax , a > 0, the corresponding rules for the function ex and thechain rule.
d
dx(ax) =
d
dx(ex ln a) = ax ln a
d
dx(ag(x)) =
d
dxeg(x) ln a = g ′(x)ag(x) ln a
I Example: Find the derivative of 5x3+2x .
I Instead of memorizing the above formulas for differentiation, I can justconvert this to an exponential function of the form eh(x) using thedefinition of 5u, where u = x3 + 2x and differentiate using the techniqueswe learned in the previous lecture.
I We have, by definition, 5x3+2x = e(x3+2x) ln 5
I Therefore ddx
5x3+2x = ddx
e(x3+2x) ln 5 = e(x3+2x) ln 5 ddx
(x3 + 2x) ln 5
I = (ln 5)(3x2 + 2)e(x3+2x) ln 5 = (ln 5)(3x2 + 2)5x3+2x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 13: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/13.jpg)
Derivatives
We can also derive the following rules of differentiation using the definition ofthe function ax , a > 0, the corresponding rules for the function ex and thechain rule.
d
dx(ax) =
d
dx(ex ln a) = ax ln a
d
dx(ag(x)) =
d
dxeg(x) ln a = g ′(x)ag(x) ln a
I Example: Find the derivative of 5x3+2x .
I Instead of memorizing the above formulas for differentiation, I can justconvert this to an exponential function of the form eh(x) using thedefinition of 5u, where u = x3 + 2x and differentiate using the techniqueswe learned in the previous lecture.
I We have, by definition, 5x3+2x = e(x3+2x) ln 5
I Therefore ddx
5x3+2x = ddx
e(x3+2x) ln 5 = e(x3+2x) ln 5 ddx
(x3 + 2x) ln 5
I = (ln 5)(3x2 + 2)e(x3+2x) ln 5 = (ln 5)(3x2 + 2)5x3+2x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 14: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/14.jpg)
Derivatives
We can also derive the following rules of differentiation using the definition ofthe function ax , a > 0, the corresponding rules for the function ex and thechain rule.
d
dx(ax) =
d
dx(ex ln a) = ax ln a
d
dx(ag(x)) =
d
dxeg(x) ln a = g ′(x)ag(x) ln a
I Example: Find the derivative of 5x3+2x .
I Instead of memorizing the above formulas for differentiation, I can justconvert this to an exponential function of the form eh(x) using thedefinition of 5u, where u = x3 + 2x and differentiate using the techniqueswe learned in the previous lecture.
I We have, by definition, 5x3+2x = e(x3+2x) ln 5
I Therefore ddx
5x3+2x = ddx
e(x3+2x) ln 5 = e(x3+2x) ln 5 ddx
(x3 + 2x) ln 5
I = (ln 5)(3x2 + 2)e(x3+2x) ln 5 = (ln 5)(3x2 + 2)5x3+2x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 15: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/15.jpg)
Graphs of General exponential functions
For a > 0 we can draw a picture of the graph of
y = ax
using the techniques of graphing developed in Calculus I.
I We get a different graph for each possible value of a.We split the analysis into two cases,
I since the family of functions y = ax slope downwards when 0 < a < 1 and
I the family of functions y = ax slope upwards when a > 1.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 16: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/16.jpg)
Graphs of General exponential functions
For a > 0 we can draw a picture of the graph of
y = ax
using the techniques of graphing developed in Calculus I.
I We get a different graph for each possible value of a.We split the analysis into two cases,
I since the family of functions y = ax slope downwards when 0 < a < 1 and
I the family of functions y = ax slope upwards when a > 1.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 17: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/17.jpg)
Graphs of General exponential functions
For a > 0 we can draw a picture of the graph of
y = ax
using the techniques of graphing developed in Calculus I.
I We get a different graph for each possible value of a.We split the analysis into two cases,
I since the family of functions y = ax slope downwards when 0 < a < 1 and
I the family of functions y = ax slope upwards when a > 1.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 18: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/18.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 19: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/19.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 20: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/20.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 21: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/21.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 22: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/22.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 23: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/23.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 24: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/24.jpg)
Case 1:Graph of y = ax , 0 < a < 1
-4 -2 2 4
10
20
30
40
50
y=1x
y=H1�8Lx
y=H1�4Lx
y=H1�2Lx
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I Slope: If 0 < a < 1, the graph ofy = ax has a negative slope andis always decreasing,ddx
(ax) = ax ln a < 0. In this casea smaller value of a gives asteeper curve [for x < 0].
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I As x →∞, x ln a approaches−∞, since ln a < 0 and thereforeax = ex ln a → 0.
I As x → −∞, x ln a approaches∞, since both x and ln a are lessthan 0. Thereforeax = ex ln a →∞.
For 0 < a < 1, limx→∞ ax = 0, lim
x→−∞ax = ∞ .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 25: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/25.jpg)
Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 26: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/26.jpg)
Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
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Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
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Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
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Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 30: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/30.jpg)
Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 31: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/31.jpg)
Case 2: Graph of y = ax , a > 1
-4 -2 2 4
20
40
60
80
100
120
y=8x
y=4x
y=2x
I y-intercept: The y-intercept isgiven byy = a0 = e0 ln a = e0 = 1.
I x-intercept: The values ofax = ex ln a are always positiveand there is no x intercept.
I If a > 1, the graph of y = ax hasa positive slope and is alwaysincreasing, d
dx(ax) = ax ln a > 0.
I The graph is concave up sincethe second derivative isd2
dx2 (ax) = ax(ln a)2 > 0.
I In this case a larger value of agives a steeper curve [whenx > 0].
I As x →∞, x ln a approaches ∞,since ln a > 0 and thereforeax = ex ln a →∞
I As x → −∞, x ln a approaches−∞, since x < 0 and ln a > 0.Therefore ax = ex ln a → 0.
For a > 1, limx→∞ ax = ∞, lim
x→−∞ax = 0 .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 32: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/32.jpg)
Power Rules
We now have 4 different types of functions involving bases and powers. So farwe have dealt with the first three types:If a and b are constants and g(x) > 0 and f (x) and g(x) are bothdifferentiable functions.
d
dxab = 0,
d
dx(f (x))b = b(f (x))b−1f ′(x),
d
dxag(x) = g ′(x)ag(x) ln a,
d
dx(f (x))g(x)
For ddx
(f (x))g(x), we use logarithmic differentiation or write the function as
(f (x))g(x) = eg(x) ln(f (x)) and use the chain rule.
Example Differentiate x2x2
, x > 0.
I Also to calculate limits of functions of this type it may help write thefunction as (f (x))g(x) = eg(x) ln(f (x)).
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 33: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/33.jpg)
Power Rules
We now have 4 different types of functions involving bases and powers. So farwe have dealt with the first three types:If a and b are constants and g(x) > 0 and f (x) and g(x) are bothdifferentiable functions.
d
dxab = 0,
d
dx(f (x))b = b(f (x))b−1f ′(x),
d
dxag(x) = g ′(x)ag(x) ln a,
d
dx(f (x))g(x)
For ddx
(f (x))g(x), we use logarithmic differentiation or write the function as
(f (x))g(x) = eg(x) ln(f (x)) and use the chain rule.
Example Differentiate x2x2
, x > 0.
I Also to calculate limits of functions of this type it may help write thefunction as (f (x))g(x) = eg(x) ln(f (x)).
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 34: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/34.jpg)
General Logarithmic Functions
Since f (x) = ax is a monotonic function whenever a 6= 1, it has an inversewhich we denote by
f −1(x) = loga x .
I We get the following from the properties of inverse functions:
I
f −1(x) = y if and only if f (y) = x
loga(x) = y if and only if ay = x
I
f (f −1(x)) = x f −1(f (x)) = x
aloga(x) = x loga(ax) = x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 35: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/35.jpg)
General Logarithmic Functions
Since f (x) = ax is a monotonic function whenever a 6= 1, it has an inversewhich we denote by
f −1(x) = loga x .
I We get the following from the properties of inverse functions:
I
f −1(x) = y if and only if f (y) = x
loga(x) = y if and only if ay = x
I
f (f −1(x)) = x f −1(f (x)) = x
aloga(x) = x loga(ax) = x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 36: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/36.jpg)
General Logarithmic Functions
Since f (x) = ax is a monotonic function whenever a 6= 1, it has an inversewhich we denote by
f −1(x) = loga x .
I We get the following from the properties of inverse functions:
I
f −1(x) = y if and only if f (y) = x
loga(x) = y if and only if ay = x
I
f (f −1(x)) = x f −1(f (x)) = x
aloga(x) = x loga(ax) = x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 37: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/37.jpg)
General Logarithmic Functions
Since f (x) = ax is a monotonic function whenever a 6= 1, it has an inversewhich we denote by
f −1(x) = loga x .
I We get the following from the properties of inverse functions:
I
f −1(x) = y if and only if f (y) = x
loga(x) = y if and only if ay = x
I
f (f −1(x)) = x f −1(f (x)) = x
aloga(x) = x loga(ax) = x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 38: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/38.jpg)
Change of base Formula
It is not difficult to show that loga x has similar properties to ln x = loge x .This follows from the Change of Base Formula which shows that Thefunction loga x is a constant multiple of ln x .
loga x =ln x
ln a
I Let y = loga x .
I Since ax is the inverse of loga x , we have ay = x .
I Taking the natural logarithm of both sides, we get y ln a = ln x ,
I which gives, y = ln xln a
.
I The algebraic properties of the natural logarithm thus extend to generallogarithms, by the change of base formula.
loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x r ) = r loga(x).
for any positive number a 6= 1. In fact for most calculations (especiallylimits, derivatives and integrals) it is advisable to convert loga x to naturallogarithms. The most commonly used logarithm functions are log10 x andln x = loge x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 39: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/39.jpg)
Change of base Formula
It is not difficult to show that loga x has similar properties to ln x = loge x .This follows from the Change of Base Formula which shows that Thefunction loga x is a constant multiple of ln x .
loga x =ln x
ln a
I Let y = loga x .
I Since ax is the inverse of loga x , we have ay = x .
I Taking the natural logarithm of both sides, we get y ln a = ln x ,
I which gives, y = ln xln a
.
I The algebraic properties of the natural logarithm thus extend to generallogarithms, by the change of base formula.
loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x r ) = r loga(x).
for any positive number a 6= 1. In fact for most calculations (especiallylimits, derivatives and integrals) it is advisable to convert loga x to naturallogarithms. The most commonly used logarithm functions are log10 x andln x = loge x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 40: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/40.jpg)
Change of base Formula
It is not difficult to show that loga x has similar properties to ln x = loge x .This follows from the Change of Base Formula which shows that Thefunction loga x is a constant multiple of ln x .
loga x =ln x
ln a
I Let y = loga x .
I Since ax is the inverse of loga x , we have ay = x .
I Taking the natural logarithm of both sides, we get y ln a = ln x ,
I which gives, y = ln xln a
.
I The algebraic properties of the natural logarithm thus extend to generallogarithms, by the change of base formula.
loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x r ) = r loga(x).
for any positive number a 6= 1. In fact for most calculations (especiallylimits, derivatives and integrals) it is advisable to convert loga x to naturallogarithms. The most commonly used logarithm functions are log10 x andln x = loge x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 41: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/41.jpg)
Change of base Formula
It is not difficult to show that loga x has similar properties to ln x = loge x .This follows from the Change of Base Formula which shows that Thefunction loga x is a constant multiple of ln x .
loga x =ln x
ln a
I Let y = loga x .
I Since ax is the inverse of loga x , we have ay = x .
I Taking the natural logarithm of both sides, we get y ln a = ln x ,
I which gives, y = ln xln a
.
I The algebraic properties of the natural logarithm thus extend to generallogarithms, by the change of base formula.
loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x r ) = r loga(x).
for any positive number a 6= 1. In fact for most calculations (especiallylimits, derivatives and integrals) it is advisable to convert loga x to naturallogarithms. The most commonly used logarithm functions are log10 x andln x = loge x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 42: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/42.jpg)
Change of base Formula
It is not difficult to show that loga x has similar properties to ln x = loge x .This follows from the Change of Base Formula which shows that Thefunction loga x is a constant multiple of ln x .
loga x =ln x
ln a
I Let y = loga x .
I Since ax is the inverse of loga x , we have ay = x .
I Taking the natural logarithm of both sides, we get y ln a = ln x ,
I which gives, y = ln xln a
.
I The algebraic properties of the natural logarithm thus extend to generallogarithms, by the change of base formula.
loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x r ) = r loga(x).
for any positive number a 6= 1. In fact for most calculations (especiallylimits, derivatives and integrals) it is advisable to convert loga x to naturallogarithms. The most commonly used logarithm functions are log10 x andln x = loge x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 43: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/43.jpg)
Change of base Formula
It is not difficult to show that loga x has similar properties to ln x = loge x .This follows from the Change of Base Formula which shows that Thefunction loga x is a constant multiple of ln x .
loga x =ln x
ln a
I Let y = loga x .
I Since ax is the inverse of loga x , we have ay = x .
I Taking the natural logarithm of both sides, we get y ln a = ln x ,
I which gives, y = ln xln a
.
I The algebraic properties of the natural logarithm thus extend to generallogarithms, by the change of base formula.
loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x r ) = r loga(x).
for any positive number a 6= 1. In fact for most calculations (especiallylimits, derivatives and integrals) it is advisable to convert loga x to naturallogarithms. The most commonly used logarithm functions are log10 x andln x = loge x .
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 44: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/44.jpg)
Using Change of base Formula for derivatives
Change of base formula
loga x =ln x
ln a
From the above change of base formula for loga x , we can easily derive thefollowing differentiation formulas:
d
dx(loga x) =
d
dx
ln x
ln a=
1
x ln a
d
dx(loga g(x)) =
g ′(x)
g(x) ln a.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 45: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/45.jpg)
A special Limit
We derive the following limit formula by taking the derivative of f (x) = ln x atx = 1, We know that f ′(1) = 1/1 = 1. We also know that
f ′(1) = limx→0
ln(1 + x)− ln 1
x= lim
x→0ln(1 + x)1/x = 1.
Applying the (continuous) exponential function to the limit on the left handside (of the last equality), we get
e limx→0 ln(1+x)1/x
= limx→0
e ln(1+x)1/x
= limx→0
(1 + x)1/x .
Applying the exponential function to the right hand sided(of the last equality),we gat e1 = e. Hence
e = limx→0
(1 + x)1/x
Note If we substitute y = 1/x in the above limit we get
e = limy→∞
“1 +
1
y
”y
and e = limn→∞
“1 +
1
n
”n
where n is an integer (see graphs below). We look at large values of n below toget an approximation of the value of e.
Annette Pilkington Natural Logarithm and Natural Exponential
![Page 46: General Logarithms and Exponentials](https://reader031.vdocuments.us/reader031/viewer/2022012021/6168a47cd394e9041f7171c6/html5/thumbnails/46.jpg)
A special Limit
n = 10→“
1 + 1n
”n
= 2.59374246, n = 100→“
1 + 1n
”n
= 2.70481383,
n = 100→“
1 + 1n
”n
= 2.71692393, n = 1000→“
1 + 1n
”n
= 2.1814593.
20 40 60 80 100
2.58
2.60
2.62
2.64
2.66
2.68
2.70
points Hn, H1 + 1�nLnL, n = 1...100
Annette Pilkington Natural Logarithm and Natural Exponential