Inverse Functions and Logarithms Calculus I PDF

Inverse Functions and Logarithms Inverse Functions Inverse Functions and Logarithms Inverse Functions Definition 1 A function f is called a one-to-one function if it never takes on the same value twice; that is, f (x1 ) 6= f (x2 ) whenever x1 6=...

Inverse Functions and Logarithms Inverse Functions Inverse Functions and Logarithms Inverse Functions Definition 1 A function f is called a one-to-one function if it never takes on the same value twice; that is, f (x1 ) 6= f (x2 ) whenever x1 6= x2. MAT 1001 Calculus I 1 / 79  ▲ In the language of inputs and outputs, Inverse Functions and Logarithms Inverse Functions 1 Definition this definition says that f is one-to- one if the same val each output corresponds to only one If a horizontal line intersects the graph of f in more than one point, then input. we see from Figure that there are numbers x1 and x2 such that f (x1 ) = f (x2 ). This means that f is not one-to-one. y If a horizont from Figure 2 t that f is not one y=ƒ mining whether fl ‡ Horizontal Line intersects its 0 ⁄ ¤ x FIGURE 2 EXAMPLE 1 Is th This function is not one-to-one MAT 1001 Calculus I SOLUTION 1 If2 / x791  Inverse Functions and Logarithms Inverse Functions Therefore, we have the following geometric method for determining whether a function is one-to-one. Horizontal Line Test A function is one-to-one if and only if no horizontal line intersects its graph more than once. MAT 1001 Calculus I 3 / 79  Inverse Functions and Logarithms Inverse Functions Example 2 Is the function f (x) = x3 one-to-one? Solution 1. If x1 6= x2 , then x31 6= x32 (two diferent numbers can’t have the same cube). Therefore, by definition, f (x) = x3 is one-to-one. MAT 1001 Calculus I 4 / 79  0 Inverse Functions and Logarithms ⁄ ¤ x Inverse Functions Solution 2. FIGURE 2 EXAMPLE 1 Thisno From Figure we see that function is not horizontal line one-to-one intersects the graph of 3 SOLUTION 1 f (x) = x more thanbecause once. Therefore, f(⁄)=f(¤).by the Horizontal Line Test, f is one-to-one. cube). Th y SOLUTION 2 y=˛ f 共x兲 苷 x 0 x EXAMPLE 2 SOLUTION 1 FIGURE 3 and so 1 MAT 1001 ƒ=˛ is one-to-one. Calculus I 5 / 79  Inverse Functions and Logarithms Inverse Functions Example 3 Is the function g(x) = x2 one-to-one? Solution 1. This function is not one-to-one because, for instance, g(1) = 1 = g(−1) and so 1 and −1 have the same output. MAT 1001 Calculus I 6 / 79  Inverse Functions and Logarithms Inverse Functions Solution 2. FIGURE 3 From Figure we see that there are horizontal lines that intersect the and graphso ƒ=˛ is one-to-one. of g more than once. Therefore, by the Horizontal Line Test, g is not one-to-one. SOLUTION y graph o y=≈ one. One possess 0 x 2 The FIGURE 4 MAT 1001 Calculus I 7 / 79  Inverse Functions and Logarithms Inverse Functions Definition 4 Let f be a one-to-one function with domain A and range B. Then its inverse function f −1 has domain B and range A and is defined by f −1 (y) = x ⇔ f (x) = y for any y in B. MAT 1001 Calculus I 8 / 79 Inverse Functions and Logarithms Inverse Functions 66 domain CHAPTER of 1 FUNCTIONS f −1 = range of f AND M −1 range of f = domain of f. x A f f –! B y FIGURE 5 MAT 1001 Calculus I 9 / 79  Inverse Functions and Logarithms Inverse Functions For example, the inverse function of f (x) = x3 is f −1 (x) = x1/3 because if y = x3 , then f −1 (y) = f −1 (x3 ) = (x3 )1/3 = x CAUTION: Do not mistake the −1 in f −1 for an exponent. Thus f −1 does not mean 1/f. MAT 1001 Calculus I 10 / 79  Inverse Functions and Logarithms Inverse Functions The letter x is traditionally used as the independent variable, so when we concentrate on f −1 rather than on f , we usually reverse the roles of x and y and write f −1 (x) = y ⇔ f (y) = x. (1) By substituting for y in Definition and substituting for x in (1), we get the following cancellation equations: f −1 (f (x)) = x x∈A f (f −1 (x)) = x x ∈ B. MAT 1001 Calculus I 11 / 79  Inverse Functions and Logarithms Inverse Functions How to Find the Inverse Function of a One-to-One Function 1 Write y = f (x). 2 Solve this equation for x in terms of y (if possible). 3 To express f −1 as a function of x, interchange x and y. The resulting equation is y = f −1 (x). MAT 1001 Calculus I 12 / 79  Inverse Functions and Logarithms Inverse Functions Example 5 Find the inverse function of f (x) = x3 + 2. Solution. According to steps in previous slide, we first write y = x3 + 2 Then we solve this equation for x: x3 = y − 2 p x= 3 y−2 Finally, we interchange x and y: √ 3 y= x − 2. √ Therefore, the inverse function is f −1 (x) = 3 x − 2. MAT 1001 Calculus I 13 / 79 8 CHAPTER 1 FUNCTIONS Inverse Functions and Logarithms AND MODELS Inverse Functions if f ⫺1共b兲 苷 a, the point 共a, b兲 is on the ⫺1 The principle of interchanging x andthe graph y to find of thef inverse. Butfunction we get also the point y 苷 x. (See Figure −1 8.) gives us the method for obtaining the graph of f from the graph of f. y (b, a) Since f (a) = b if and only if f −1 (b) = a, the point (a, b) is on the graph of f if and only if the point (b, a) is on the graph (a, b) of f −1. But we get the point 0 (b, a) from by reflecting about x the line y = x. y=x FIGURE 8 MAT 1001 Therefore, as illustrated by Figure Calculus I 14 / 799:  Inverse Functions and Logarithms Inverse Functions The graph of f −1 is obtained by reflecting the graph of f about the line y = x. y 0 x MAT 1001 Calculus I 15 / 79  Inverse Functions and Logarithms Inverse Functions The graph of f −1 is obtained by reflecting the graph of f about the line y = x. y 0 x MAT 1001 Calculus I 15 / 79  Inverse Functions and Logarithms Inverse Functions The graph of f −1 is obtained by reflecting the graph of f about the line y = x. y 0 x MAT 1001 Calculus I 15 / 79  Inverse Functions and Logarithms Inverse Functions Example 6 √ 0 Sketch the graphs of f (x) = −1 − x and its inverse function using the same coordinate axes. y=x Solution. √ First we sketch the curve y = −1 − x (the top half of the parabola FIGURE 8 y 2 = −1 − x, , or x = −y 2 − 1) and then we reflect about the line y = x −1 to get the graph of f. Therefore, as illustrate y y=ƒ The graph of f ⫺1 is ob y=x 0 EXAMPLE 5 Sketch the gra (_1, 0) x same coordinate axes. (0, _1) SOLUTION First we sketch th y=f –!(x) y 2 苷 ⫺1 ⫺ x, or x 苷 ⫺y graph of f ⫺1. (See Figure for f ⫺1 is f ⫺1共x兲 苷 ⫺x 2 ⫺ MAT 1001 FIGURE 10 Calculus I parabola y 苷 ⫺x 2 ⫺ 16 /179an  Inverse Functions and Logarithms Inverse Functions Solution (cont.) As a check on our graph, notice that the expression for f −1 isFIGURE 8 f −1 (x) = −x2 − 1, x > 0. So the graph of f −1 is the right half of the Therefore parabola y = −x2 − 1 and this seems reasonable from Figure. y y=ƒ The graph y=x 0 EXAMPLE 5 S (_1, 0) x same coordin (0, _1) SOLUTION First y=f –!(x) y 2 苷 ⫺1 ⫺ graph of f ⫺1 for f ⫺1 is f ⫺ MAT 1001 FIGURE 10 Calculus I parabola17y/ 79苷  Trigonometry Angles Trigonometry Angles Angles can be measured in degrees or in radians (abbreviated as rad). The angle given by a complete revolution contains 360◦ , which is the same as 2π rad. Therefore π rad = 180◦ (2) and  ◦ 180 π 1 rad = ≈ 57.3◦ 1◦ = rad ≈ 0.017 rad (3) π 180 MAT 1001 Calculus I 18 / 79  Trigonometry Angles Example 7 (a) Find the radian measure of 60◦. (b) Express 5π/4 rad in degrees. Solution. (a) From Equation (2) or (3) we see that to convert from degrees to radians we multiply by π/180. Therefore  π  π 60◦ = 60 = rad 180 3 (b) To convert from radians to degrees we multiply by 180/π. Thus   5π 5π 180 rad = = 225◦ 4 4 π MAT 1001 Calculus I 19 / 79  Trigonometry Angles In calculus we use radians to measure angles except when otherwise indicated. The following table gives the correspondence between degree and radian measures of some common angles. Degrees 0◦ 30◦ 45◦ 60◦ 90◦ 120◦ 135◦ π π π π 2π 3π Radians 0 6 4 3 2 3 4 Degrees 150◦ 180◦ 270◦ 360◦ 5π 3π Radians 6 π 2 2π MAT 1001 Calculus I 20 / 79 angle of(b)3With 兾8 rad?
3 cm and 
3兾8 rad, the arc length is r Trigonometry Angles SOLUTION The standard (a) Using origin of aEquation position ofa
anrangle a
6 and 3 withsystem coordinate and rits 冉 冊 3
3 occurs
9when we place its vertex at the
initial 8 cm 5, we8see sidethat onthe theangle is positive x−axis as in FigureThe1.standard position of an angle occurs when we place its vertex at the origin of 

1.2 rad 6 5 the positive x-axis as in Figure 3. a coordinate system and its initial side on Figure 1: θ > 0 (b) With r
3 cm yand 
3兾8 rad, the arc A length y is positive angle is obtained 冉 冊 initial side 3by rotating 9 the initial side a
r
3
0 cm counterclockwise x 8 terminal sideuntil it coin- ¨ terminal 8 side ¨ initial side cides with the terminal side. The standard position 0 of an angle x occurs when we place its vertex at the origin of a coordinate system and its initial side on the positive x-axis as in Figure 3. FIGURE 3 FIGURE 4 Figure 2: θ < 0 ¨ ˘0 ¨0 cos ¨>0 P(1, s3 Find the exact trigonometric ratios for θ = 2π/3. FIGURE 9 Solution. y in the defi P {_1, œ„ 3} 2 3 œ„ 2π π 3 3 1 0 x The fo FIGURE 10 MAT 1001 Calculus I Example 27 / 79  Trigonometry The Trigonometric Functions Solution (cont.) From Figure √ we see that a point on the terminal line for θ = 2π/3 is P (−1, 3). Therefore, taking √ x = −1 y = 3 r = 2 in the definitions of the trigonometric ratios, we have √ 2π 3 2π 1 2π √ sin = cos =− tan =− 3 3 2 3 2 3 2π 2 2π 2π 1 csc =√ sec = −2 cot = −√ 3 3 3 3 3 MAT 1001 Calculus I 28 / 79  Trigonometry The Trigonometric Functions Note If θ is a number, the convention is that sin θ means the sine of the angle whose radian measure is θ. For example, the expression sin 3 implies that we are dealing with an angle of 3 rad. When finding a calculator approximation to this number we must remember to set our calculator in radian mode, and then we obtain sin 3 ≈ 0.14112 If we want to know the sine of the angle 3◦ we would write sin 3◦ and, with our calculator in degree mode, we find that sin 3◦ ≈ 0.05234 MAT 1001 Calculus I 29 / 79  sin
cos
 tan
s3 3 Trigonometry 2 3 2 3 The Trigonometric Functions 2 2 2 2 1 csc
sec
2 cot
 3 s3 3 3 s3 x The following table gives some values of sin  and cos  found by the method of The following Example 3. table gives some values of sin θ and cos θ.     2 3 5 3  0  2 6 4 3 2 3 4 6 2 1 1 s3 s3 1 1 sin  0 1 0 1 0 2 s2 2 2 s2 2 s3 1 1 1 1 s3 cos  1 0    1 0 1 2 s2 2 2 s2 2 EXAMPLE 4 If cos 
and 0

兾2, find the other five trigonometric func- 2 5 tions of . SOLUTION Since cos 
5 , we can label the hypotenuse as having length 5 and the 2 adjacent side as having length 2 in Figure 11. If the opposite side has length x, then the Pythagorean Theorem gives x 2  4
25 and so x 2
21, or x
s21. We can now use the diagram to write the other five trigonometric functions: MAT 1001 Calculus I 30 / 79  Trigonometry Trigonometric Identities Trigonometric Identities A trigonometric identity is a relationship among the trigonometric functions. The most elementary are the following, which are immediate consequences of the definitions of the trigonometric functions. 1 1 1 csc θ = sec θ = cot θ = sin θ cos θ tan θ sin θ cos θ tan θ = cot θ = cos θ sin θ MAT 1001 Calculus I 31 / 79  Trigonometry Trigonometric Identities sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ 1 + cot2 θ = csc2 θ sin(−θ) = − sin θ cos(−θ) = cos θ sin(θ + 2π) = sin θ cos(θ + 2π) = cos θ MAT 1001 Calculus I 32 / 79  Trigonometry Trigonometric Identities The remaining trigonometric identities are all consequences of two basic identities called the addition formulas: sin(α + β) = sin α cos β + cos α sin β cos(α + β) = cos α cos β − sin α sin β. MAT 1001 Calculus I 33 / 79  Trigonometry Trigonometric Identities sin(α − β) = sin α cos β − cos α sin β cos(α − β) = cos α cos β + sin α sin β tan α + tan β tan(α + β) = 1 − tan α tan β tan α − tan β tan(α − β) = 1 + tan α tan β MAT 1001 Calculus I 34 / 79  Trigonometry Trigonometric Identities sin 2α = 2 sin α cos α cos 2α = cos2 α − sin2 α cos 2α = 2 cos2 α − 1 cos2α = 1 − 2 sin2 α 1 + cos 2α cos2 α = 2 1 − cos 2α sin2 α = 2 MAT 1001 Calculus I 35 / 79  Trigonometry Trigonometric Identities Example 9 Find all values of x in the interval [0, 2π] such that sin x = sin 2x Solution. Using the double-angle formula, we rewrite the given equation as sin x = 2 sin x cos x or sin x(1 − 2 cos x) = 0 Therefore, there are two possibilities: sin x = 0 or 1 − 2 cos x = 0 1 π 5π x = 0, π, 2π or cos x = ⇒x= , 2 3 3 The given equation has five solutions: x = 0, π, 2π, π3 , 5π 3. MAT 1001 Calculus I 36 / 79  Trigonometry Graphs of the Trigonometric Functions Graphs of the Trigonometric Functions The graph of the function f (x) = sin x is obtained by plotting points for 0 ≤ x ≤ 2π and then using the periodic nature of the function to complete the graph. Notice that the zeros of the sine function occur at the integer multiples of π, that is, sin x = 0 whenever x = nπ, n an integer Because of the identity  π cos x = sin x + 2 the graph of cosine is obtained by shifting the graph of sine by an amount π/2 to the left. Note that for both the sine and cosine functions the domain is −∞, ∞ and the range is the closed interval [−1, 1]. Thus, for all values of x, we have −1 ≤ sin x ≤ 1 − 1 ≤ cos x ≤ 1. MAT 1001 Calculus I 37 / 79  the domain is 共, 兲 andGraphs ine functions Trigonometry theof therange is the closed Trigonometric Functions s, for all values of x, we y have π 1 3π _ 2 2 1  sin x  1 0 1π  cos x  1 x _π π 2π 5π 3π _1 2 2 y 1 (a) ƒ=sin x _π π 3π FIGURE 13 π 0 π 3π 2π 5π x _ 2 _1 2 2 2 The graphs of the re and their domains are i (b) ©=cos x 共, 兲are The graphs of the remaining four trigonometric functions , whereas shown in cosec following figures: tions are periodic: tange MAT 1001 Calculus I have period 2. 38 / 79  Trigonometry Graphs of the Trigonometric Functions y 1 _π 0 π π π 3π x _π _ _ 2 _1 2 2 (a) y=tan x Figure 3: y = tan x y MAT 1001 Calculus I 39 / 79  Trigonometry Graphs of the Trigonometric Functions y 3π x _π π 0 π π 3π x _ 2 2 2 2 (b) y=cot x Figure 4: y = cot x y MAT 1001 Calculus I 40 / 79  Trigonometry Graphs of the Trigonometric Functions (a) y=tan x y y=sin x _ π 1 3π 2 0 2 _π _ π π x _1 2 FIGURE 14 (c) y=csc x Figure 5: y = csc x MAT 1001 Inverse Trigonometric Calculus I Functions 41 / 79  Trigonometry Graphs of the Trigonometric Functions (b) y=cot x y sin x y=cos x 1 π π 3π 2 _π _ 2 0 2 x π π x _1 2 (d) y=sec x Figure 6: y = sec x tric Functions MAT 1001 Calculus I 42 / 79  Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Functions When we try to find the inverse trigonometric functions, we have a slight difficulty. Because the trigonometric functions are not one-to-one, they don’t have inverse functions. The difficulty is overcome by restricting the domains of these functions so that they become one-to-one. MAT 1001 Calculus I 43 / 79  Trigonometry they become one-to-one. Inverse Trigonometric Functions You can see from Figure 15 th the Horizontal Line Test). But Figure 16), is one-to-one. The inv You can see from Figure that the sine function and isy denoted = sin(x)by sin 1 or arcsin. is not one-to-one (use the Horizontal Line Test).function. y y=sin x _π 0 π π x 2 FIGURE 15 Since the definition of an inve f 1共x MAT 1001 Calculus I 44 / 79 1 Trigonometry Inverse Trigonometric Functions or arcsin. It is called the inverse sine function or the arcsine But the function f (x) = sin(x), −π/2 ≤ x ≤ π/2, is one-to-one. y _ π2 x 0 π x 2 FIGURE 16 Figure 7: n of an inverse function says that The inverse function of this restricted sine function f exists and is denoted by sin−1 1or arcsin. f 共x兲
y &? f 共y兲
x It is called the inverse sine function or the arcsine function. MAT 1001 Calculus I 45 / 79  Trigonometry Inverse Trigonometric Functions Since the definition of an inverse function says that f −1 (x) = y ⇔ f (y) = x we have sin−1 (x) = y ⇔ sin(y) = x and − π/2 ≤ y ≤ π/2 Thus, if −1 ≤ x ≤ 1, sin−1 x is the number between −π/2 and π/2 whose sine is x. MAT 1001 Calculus I 46 / 79  Trigonometry Inverse Trigonometric Functions A26 APPENDIX C TRIGONOMETRY Example 10 we have Evaluate (a) sin−1 (1/2) and (b) tan(arcsin(1/3)).   sin1x
y &? sin y
x and  y 2 2 Solution. | sin 1x  1 Thus, if 1  x  1, sin 1x is the number between 兾2 and 兾2 whose sin x (a) Since sin(π/6) = 1/2, and EXAMPLE π/6 is7 Evaluate in between −π/2 and π/2, we have (a) sin ( ) and (b) tan(arcsin ). 1 1 1 2 3 SOLUTION −1have π sin (a) We (1/2) =. 6  sin1( 12)
6 Let θsin共 because =兾6兲 arcsin(1/3). Thenwe
12 and 兾6 lies between can 兾2 and 兾2.draw 3 a right (b) triangle Let 
arcsin 3. Then with 1 angle we can draw a rightθtriangle as in withFig- angle  as in and deduce from the Pythagorean Theorem that the third side has length (b) 1 ure and deduce from the Pythagorean s9  1
2s2. This enables us to read from the triangle that ¨ 2 œ„ 2 Theorem √ that √ the third side has1 length tan(arcsin )
tan 
1 FIGURE 17 9 − 1 = 2 2. 2 s2 3 This enables us to read fromThe thecancellation triangle that equations for inverse functions [see (1.6.4)] become, i 1 1   tan(arcsin(1/3)) = tan θsin= 共sin√x兲
. x for  2 x 2 2 2 sin共sin1x兲
x for 1  x  1 MAT 1001 Calculus I 47 / 79  The inverse sine function, Trigonometry sin , has domain 关 Inverse Trigonometric Functions graph, shown in Figure 18, is obtained from tha The inverse sine function, sin−1 , has domain [−1, 1] and range ure 16) [−π/2, π/2], and its graph, by in shown reflection Figure 8, about is obtainedthe line fromythat
xof. the restricted sine function (Figure 7) by reflection about the line y = x. y π 2 _1 0 1 x _ π2 _ π2 Figure 8: MAT 1001 FIGURE 18 Calculus y=sin–! I x FIGURE 4819 / 79 y  1 , has domain 关1, 1兴 and range ine function, sinTrigonometry 关 Inverse 兾2, Functions Trigonometric 兾2兴, and its inThe Figure 18, is tangent obtained function canfrom that one-to-one be made of the restricted sine function by restricting (Fig- it to the flection about interval line y
x. the π/2). (−π/2, y 0 1 x _ π2 0 π x 2 _ π2 =sin–! x y=tan9:x, _ π2 0 and a 6= 1, the exponential function f (x) = ax is either increasing or decreasing and so it is one-to-one by the Horizontal Line Test. It therefore has an inverse function f −1 , which is called the logarithmic function with base a and is denoted by loga. MAT 1001 Calculus I 64 / 79  Exponential Functions Logarithmic Functions If we use the formulation of an inverse function f −1 (x) = y ⇐⇒ f (y) = x then we have loga x = y ⇐⇒ ay = x. Thus, if 0 < x, then loga x, a is the exponent to which the base must be raised to give x. For example, log10 0.001 = −3 because 10−3 = 0, 001. MAT 1001 Calculus I 65 / 79  Exponential Functions Logarithmic Functions The cancellation equations, when applied to f (x) = ax and f −1 (x) = loga x become loga (ax ) = x , x∈R aloga x = x , x > 0. In particular, if we set x = 1, we get loga (a) = 1. MAT 1001 Calculus I 66 / 79  Exponential Functions Logarithmic Functions The logarithmic function loga x has domain (0, ∞) and range R. Its graph is the reflection of the graph of y = ax about the line y = x. y The loga y=x reflection o Figure 1 have base a is reflected y=a®, a>1 Figure 1 log a 1 苷 0, 0 x The foll y=log a x, a>1 properties o Laws of L Figure shows the case where 1 < a. (The most important logarithmic FIGURE 11 functions have base a > 1.) 1. log 共x a MAT 1001 Calculus I 67 / 79  Exponential Functions Logarithmic Functions y The fact that y = ax is a very rapidly increasing function for 0 < x is re- flected in the fact that y = x loga x is a very slowly in- creasing function for 1 < x. Figure shows the graphs of y = loga x with various values of the base a. Since loga 1 = 0, the graphs of all logarithmic functions pass through the point (1, 0). MAT 1001 Calculus I 68 / 79  Exponential Functions Logarithmic Functions y The fact that y = ax is a very rapidly increasing function for 0 < x is re- flected in the fact that y = x loga x is a very slowly in- creasing function for 1 < x. Figure shows the graphs of y = loga x with various values of the base a. Since loga 1 = 0, the graphs of all logarithmic functions pass through the point (1, 0). MAT 1001 Calculus I 68 / 79  Exponential Functions Logarithmic Functions y log2 x log5 x The fact that y = ax is a very rapidly increasing log10 x function for 0 < x is re- flected in the fact that y = 1 x loga x is a very slowly in- creasing function for 1 < x. Figure shows the graphs of y = loga x with various values of the base a. Since loga 1 = 0, the graphs of all logarithmic functions pass through the point (1, 0). MAT 1001 Calculus I 68 / 79  Exponential Functions Logarithmic Functions Laws of Logarithms If x and y are positive numbers, then 1 loga (xy) = loga x + loga y   x 2 loga = loga x − loga y y 3 loga (xr ) = r loga x (where r is any real number.) MAT 1001 Calculus I 69 / 79  Exponential Functions Logarithmic Functions Example 14 Use the laws of logarithms to evaluate log2 80 − log2 5. Solution. Using Law 2, we have   80 log2 80 − log2 5 = log2 = log2 16 = 4 5 because 24 = 16. MAT 1001 Calculus I 70 / 79  Exponential Functions Natural Logarithm Natural Logarithm The logarithm with base e is called the natural logarithm and has a special notation: loge x = ln x The defining properties of the natural logarithm function are ln x = y ⇐⇒ ey = x (6) ln(ex ) = x x∈R eln x = x x > 0. (7) MAT 1001 Calculus I 71 / 79  Exponential Functions Natural Logarithm In particular, if we set x = 1, we get ln e = 1. For any positive number a, we have ln x loga x = , a > 0, a 6= 1. ln a MAT 1001 Calculus I 72 / 79  Exponential Functions Natural Logarithm Example 15 Find x if ln x = 5. Solution 1. From equation (6) we see that ln x = 5 means e5 = x Therefore, x = e5. MAT 1001 Calculus I 73 / 79  Exponential Functions Natural Logarithm Solution 2. Start with the equation ln x = 5 and apply the exponential function to both sides of the equation: eln x = e5 But the second cancellation in equation (7) says that eln x = x. Therefore, x = e5. MAT 1001 Calculus I 74 / 79  Exponential Functions Natural Logarithm Example 16 Solve the equation e5−3x = 10. Solution. We take natural logarithms of both sides of the equation and use equation (7): ln(e5−3x ) = ln 10 5 − 3x = ln 10 3x = 5 − ln 10 1 x = (5 − ln 10) 3 Since the natural logarithm is found on scientific calculators, we can approximate the solution to four decimal places: x ≈ 0.8991. MAT 1001 Calculus I 75 / 79  Exponential Functions Natural Logarithm Example 17 1 Express ln a + ln b as a single logarithm. 2 Solution. Using Laws 3 and 1 of logarithms, we have 1 ln a + ln b = ln a + ln b1/2 2 √ = ln a + ln b √ = ln(a b). MAT 1001 Calculus I 76 / 79  Exponential Functions So the inverse fu Natural Logarithm The graphs of the exponential function y = ex and its inverse function, the natural logarithm function, are shown in Figure 10. Because the curve y = ex crosses the y−axis with a slope of 1, it follows that the reflected curve y = ln x crosses the x-axis with a slope of 1. y This function giv y=´ ticular, the time r y=x 1 y=ln x This answer agre 0 tion 1.5. 1 x The graphs of logarithm functio y-axis with a slop FIGURE 13 Figure 10: with a slope of 1 MAT 1001 Calculus I 77 / 79  Exponential Functions Natural Logarithm Example 18 Sketch the graph of the function y = ln(x − 2) − 1. Solution. y lnx We start with the graph of y = ln x as given in Fig- ure 10. Using the trans- formations, we shift it two units to the right to get 0 x the graph of y = ln(x − 2) and then we shift it one unit downward to get the graph of y = ln(x − 2) − 1. MAT 1001 Calculus I 78 / 79  Exponential Functions Natural Logarithm Example 18 Sketch the graph of the function y = ln(x − 2) − 1. Solution. y lnx We start with the graph of y = ln x as given in Fig- ure 10. Using the trans- formations, we shift it two units to the right to get 0 x the graph of y = ln(x − 2) and then we shift it one unit downward to get the graph of y = ln(x − 2) − 1. MAT 1001 Calculus I 78 / 79  Exponential Functions Natural Logarithm Example 18 Sketch the graph of the function y = ln(x − 2) − 1. Solution. y lnx ln(x-2) We start with the graph of y = ln x as given in Fig- ure 10. Using the trans- ln(x-2)-1 formations, we shift it two units to the right to get 0 2 3 x the graph of y = ln(x − 2) and then we shift it one unit downward to get the -1 graph of y = ln(x − 2) − 1. MAT 1001 Calculus I 78 / 79  Exponential Functions Natural Logarithm Although ln x is an increasing function, it grows very slowly when 1 < x. In fact, ln x grows more slowly than any positive power of x. To illustrate this fact, we compare the graphs of the functions y = ln x and √ y = x1/2 = x in Figure. You can see that initially the graphs of and grow at comparable rates, but eventually the root function far surpasses the logarithm. y 2.0 1.5 x 1.0 ln x 0.5 0.0 x 0.5 1.0 1.5 2.0 2.5 3.0 -0.5 MAT 1001 Calculus I 79 / 79  Exponential Functions Natural Logarithm Although ln x is an increasing function, it grows very slowly when 1 < x. In fact, ln x grows more slowly than any positive power of x. To illustrate this fact, we compare the graphs of the functions y = ln x and √ y = x1/2 = x in Figure. You can see that initially the graphs of and grow at comparable rates, but eventually the root function far surpasses the logarithm. y 2.0 1.5 x 1.0 ln x 0.5 0.0 x 0.5 1.0 1.5 2.0 2.5 3.0 -0.5 MAT 1001 Calculus I 79 / 79  Exponential Functions Natural Logarithm Although ln x is an increasing function, it grows very slowly when 1 < x. In fact, ln x grows more slowly than any positive power of x. To illustrate this fact, we compare the graphs of the functions y = ln x and √ y = x1/2 = x in Figure. You can see that initially the graphs of and grow at comparable rates, but eventually the root function far surpasses the logarithm. y 25 20 x 15 ln x 10 5 x 200 400 600 800 MAT 1001 Calculus I 79 / 79