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Home , Logarithm

General and Exponentials

Last day, we looked at the inverse of the , the . we have the following :

ln(x) ex

a x ln(x) ln(ab) = ln a+ln b, ln( ) = ln a−ln b ln e = x and e = x b ln ax = x ln a ex ex+y = ex ey , ex−y = , (ex )y = exy . ey

lim ln x = ∞, lim ln x = −∞ x x x→∞ x→0 lim e = ∞, and lim e = 0 x→∞ x→−∞ d 1 d ln |x| = ex = ex dx x dx Z 1 Z dx = ln |x| + ex dx = ex + C x

Annette Pilkington and Natural Exponential General exponential functions

For a > 0 and x any , we define

ax = ex ln a, a > 0.

The function ax is called the exponential function with a. Note that ln(ax ) = x ln a is true for all real numbers x and all a > 0. (We saw this before for x a ). Note: The above definition for ax does not apply if a < 0.

Annette Pilkington Natural Logarithm and Natural Exponential Laws of Exponents

We can derive the following laws of exponents directly from the definition and the corresponding laws for the exponential function ex :

ax ax+y = ax ay ax−y = (ax )y = axy (ab)x = ax bx ay

I For example, we can prove the first rule in the following way: x+y (x+y) ln a I a = e x ln a+y ln a I = e x ln a y ln a x y I = e e = a a .

I The other laws follow in a similar manner.

Annette Pilkington Natural Logarithm and Natural Exponential

We can also derive the following rules of differentiation using the definition of the function ax , a > 0, the corresponding rules for the function ex and the . d d d d (ax ) = (ex ln a) = ax ln a (ag(x)) = eg(x) ln a = g 0(x)ag(x) ln a dx dx dx dx

x3+2x I Example: Find the of 5 .

I Instead of memorizing the above formulas for differentiation, I can just convert this to an exponential function of the form eh(x) using the definition of 5u, where u = x 3 + 2x and differentiate using the techniques we learned in the previous lecture. x3+2x (x3+2x) ln 5 I We have, by definition, 5 = e d x3+2x d (x3+2x) ln 5 (x3+2x) ln 5 d 3 I Therefore dx 5 = dx e = e dx (x + 2x) ln 5 2 (x3+2x) ln 5 2 x3+2x I = (ln 5)(3x + 2)e = (ln 5)(3x + 2)5 .

Annette Pilkington Natural Logarithm and Natural Exponential 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 I.

I We get a different graph for each possible value of a. We split the analysis into two cases, x I since the family of functions y = a downwards when 0 < a < 1 and x I the family of functions y = a slope upwards when a > 1.

Annette Pilkington Natural Logarithm and Natural Exponential Case 1:Graph of y = ax , 0 < a < 1

Slope: If 0 < a < 1, the graph of y=H12Lx I y = ax has a negative slope and 50 y=H14Lx is always decreasing, d (ax ) = ax ln a < 0. In this case 40 y=H18Lx dx a smaller value of a gives a

30 y=1x steeper curve [for x < 0].

I The graph is concave up since 20 the second derivative is 2 d (ax ) = ax (ln a)2 > 0. 10 dx2 I As x → ∞, x ln a approaches -4 -2 2 4 −∞, since ln a < 0 and therefore ax = ex ln a → 0. I y-intercept: The y-intercept is I As x → −∞, x ln a approaches given by ∞, since both x and ln a are less y = a0 = e0 ln a = e0 = 1. than 0. Therefore ax = ex ln a → ∞. I x-intercept: The values of x x x x ln a For 0 < a < 1, lim a = 0, lim a = ∞ . a = e are always positive x→∞ x→−∞ and there is no x intercept.

Annette Pilkington Natural Logarithm and Natural Exponential Case 2: Graph of y = ax , a > 1

x I If a > 1, the graph of y = a has = x y 2 a positive slope and is always 120 d x x increasing, dx (a ) = a ln a > 0. y=4x 100 I The graph is concave up since

y=8x 80 the second derivative is d2 x x 2 dx2 (a ) = a (ln a) > 0. 60 I In this case a larger value of a 40 gives a steeper curve [when

20 x > 0].

I As x → ∞, x ln a approaches ∞, -4 -2 2 4 since ln a > 0 and therefore ax = ex ln a → ∞ I y-intercept: The y-intercept is given by I As x → −∞, x ln a approaches −∞, since x < 0 and ln a > 0. y = a0 = e0 ln a = e0 = 1. Therefore ax = ex ln a → 0. I x-intercept: The values of x x x x ln a For a > 1, lim a = ∞, lim a = 0 . a = e are always positive x→∞ x→−∞ and there is no x intercept.

Annette Pilkington Natural Logarithm and Natural Exponential Rules

We now have 4 different types of functions involving bases and powers. So far we have dealt with the first three types: If a and b are constants and g(x) > 0 and f (x) and g(x) are both differentiable functions. d d d ab = 0, (f (x))b = b(f (x))b−1f 0(x), ag(x) = g 0(x)ag(x) ln a, dx dx dx d (f (x))g(x) dx d g(x) For dx (f (x)) , we use logarithmic differentiation or write the function as (f (x))g(x) = eg(x) ln(f (x)) and use the chain rule.

I Also to calculate limits of functions of this type it may help write the function as (f (x))g(x) = eg(x) ln(f (x)).

Annette Pilkington Natural Logarithm and Natural Exponential Example

2 Example Differentiate x 2x , x > 0. 2x2 I We use logarithmic differentiation on y = x .

I Applying the natural logarithm to both sides, we get

ln(y) = 2x 2 ln(x)

I Differentiating both sides, we get 1 dy 2x 2 = (ln x)4x + . y dx x

dy h i 2x2 h i I Therefore dx = y 4x ln x + 2x = x 4x ln x + 2x .

Annette Pilkington Natural Logarithm and Natural Exponential Example

Example What is lim x −x x→∞

−x −x ln(x) I limx→∞ x = limx→∞ e

I As x → ∞, we have x → ∞ and ln(x) → ∞, therefore if we let u = −x ln(x), we have that u approaches −∞ as x → ∞.

I Therefore lim e−x ln(x) = lim eu = 0 x→∞ u→−∞

Annette Pilkington Natural Logarithm and Natural Exponential General Logarithmic Functions

Since f (x) = ax is a whenever a 6= 1, it has an inverse which we denote by −1 f (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

y loga(x) = y if and only if a = x

I f (f −1(x)) = x f −1(f (x)) = x

loga(x) x a = x loga(a ) = x.

Annette Pilkington Natural Logarithm and Natural Exponential Change of base

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 The

function loga x is a constant multiple of ln x. ln x log x = a ln a

I Let y = loga x. x y I Since a is the inverse of loga x, we have a = x. I Taking the natural logarithm of both sides, we get y ln a = ln x, ln x I which gives, y = ln a . I The algebraic properties of the natural logarithm thus extend to general logarithms, by the change of base formula.

r loga 1 = 0, loga(xy) = loga(x) + loga(y), loga(x ) = r loga(x). for any positive number a 6= 1. In fact for most (especially

limits, derivatives and ) it is advisable to convert loga x to natural logarithms. The most commonly used logarithm functions are log10 x and ln x = loge x. Annette Pilkington Natural Logarithm and Natural Exponential Using Change of base Formula for derivatives

Change of base formula ln x log x = a ln a

From the above change of base formula for loga x, we can easily derive the following differentiation formulas:

d d ln x 1 d g 0(x) (log x) = = (log g(x)) = . dx a dx ln a x ln a dx a g(x) ln a

Annette Pilkington Natural Logarithm and Natural Exponential A special

We derive the following limit formula by taking the derivative of f (x) = ln x at x = 1, We know that f 0(1) = 1/1 = 1. We also know that ln(1 + x) − ln 1 f 0(1) = lim = lim ln(1 + x)1/x = 1. x→0 x x→0 Applying the (continuous) exponential function to the limit on the left hand side (of the last ), we get

1/x 1/x elimx→0 ln(1+x) = lim eln(1+x) = lim(1 + x)1/x . x→0 x→0 Applying the exponential function to the right hand sided(of the last equality), we gat e1 = e. Hence e = lim(1 + x)1/x x→0 Note If we substitute y = 1/x in the above limit we get

“ 1 ”y “ 1 ”n e = lim 1 + and e = lim 1 + y→∞ y n→∞ n

where n is an (see graphs below). We look at large values of n below to get an of the value of e.

Annette Pilkington Natural Logarithm and Natural Exponential A special Limit

n n “ 1 ” “ 1 ” n = 10 → 1 + n = 2.59374246, n = 100 → 1 + n = 2.70481383,

n n “ 1 ” “ 1 ” n = 100 → 1 + n = 2.71692393, n = 1000 → 1 + n = 2.71814593.

points Hn, H1 + 1nLnL, n = 1...100 2.70

2.68

2.66

2.64

2.62

2.60

2.58

20 40 60 80 100

Annette Pilkington Natural Logarithm and Natural Exponential