Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 49 potx

Acidity is determined by the concentration of hydrogen ions in a solution.. a If the concentration of hydrogen ions in a solution is increased tenfold, what happens to the pH?. THEME AND

26 ex

− 2 =e3x

27 3ln x= 5x

28 ln(x+1)4 + 5 = 13

29. ln(2x+1)3 2− 1 = 10

(e x +1)2 = 27

31 (5π )πx+2 + π = 3

32 ln(x − 3) − ln(2x + 1) = 1

In Problems 33 throuogh 36, solve for x; Q, R, and S are positive constants.

33 (a) 35x+2= 100

35 (a) (Q + R)x

= S (b) (QR)x

= S

(b) RRx − S = R

37 Acidity is determined by the concentration of hydrogen ions in a solution The pH scale, proposed by Sorensen in the early 1900s, defines pH to be − log[H+], where [H+] is the concentration of hydrogen ions given in moles per liter A pH of 7 is considered neutral; a pH greater than 7 means the solution is basic, while a pH of less than 7 indicates acidity

(a) If the concentration of hydrogen ions in a solution is increased tenfold, what happens to the pH?

(b) If a blood sample has a hydrogen ion concentration of 3.15 x 10−8, what is the pH?

(c) You’ll find that the blood sample described in part (b) is mildly basic Which has

a higher concentration of hydrogen ions: the blood sample or something neutral? How many times greater is it?

In Problems 38 through 44 find all x for which each equation is true.

38 [log x]3= log(x3)

39 ex3

= (ex)3

40 ln x =ln x

41.ln xln 2= ln x − ln 2

42 ex+1= ex+ e1

43 102x= 10210x

ln x =12ln x

45 Suppose $M0is put in a bank account where it grows according to:

M(t ) = M0

12

12t

, where t is in years

(a) If r = 0.05, how long will it take for the amount of money in the account to increase

by 50%?

(b) If the money doubles in exactly 8 years, what is r?

46 Find the equation of the straight line through the points (2, ln 2) and (3, ln 3)

47 Find the equation of the line through the points (2, ln 2) and (2 + !, ln(2 + !))

48 The “Rule of 70” says that if a quantity grows exponentially at a rate of r% per unit of time, then its doubling time is usually about 70/r This is merely a rule of thumb Now

we will determine how accurate an estimate this is and for what values of r it should

be applied

Suppose that a quantity Q grows exponentially at r% per unit of time t Thus, Q(t ) = Q01 + r

100

t

(a) Let D(r) be the doubling time of Q as a function of r Find an equation for D(r) (b) On your graphing calculator, graph D(r) and 70/r Take note of the values of r for which the latter is a good approximation of the former

THEME AND VARIATIONS

In your mind’s eye you should carry a picture of a pair of simple exponential and logarithmic functions (like 10xand log x, for instance), because a picture tells you a lot about how the functions behave “Why clutter my mind? I can always consult my graphing calculator,” you might be thinking That’s a bit like saying, “Why remember my mother’s and my father’s names? I can always ask my sister; she knows.” The exponential and logarithmic functions behave so very differently that you want to be able to have identifiers for them It’s not difficult to reconstruct an exponential graph for yourself if necessary, and from that you can construct a logarithmic graph (If you want to get a feel for ex, try 3x.)

x

–3 –2 –1 1

1 2

2

3

3

4 –1

f (x) = e x

y

x

–2 –1 1

1 2

2

3

3

4 –1

f (x) = ln x

Figure 13.4

We briefly recap some important characteristics of the exponential and logarithmic functions

Exponentials with a base greater than 1 grow increasingly rapidly A log graph grows increasingly slowly, although the log function grows without bound: As x → ∞, log x → ∞ Think of y = log10x for a moment If x = 10, then y = 1 To get to a height of 2, x must be 100 To reach a height of 3, x must increase to 1000; for a height

of 6, x must be 1 million This is sluggish growth

Exponential functions are defined for all real numbers, but the range of bx is only positive numbers On the other hand, logbxis defined only for positive numbers, while its range is all real numbers These are not functions you want to confuse

In this section we will play around a bit with some fancier variations on the basic logarithmic function and its graph

EXAMPLE 13.11 Sketch f (x) = − log2x Label at least three points on the graph

SOLUTION log2xand 2xare inverse functions We are familiar with the graph of 2x, so we’ll begin with

the graph of y = 2xand reflect it over the line y = x (interchanging the x- and y-coordinates

of the points) to obtain the graph of y = log2x

y

x

–1 1

1

2 2

–1 –2

y = 2x

y = log2 x

(.5, –1) (–1, 5) (2, 1)

(1, 2)

Figure 13.5

To obtain the graph of f (x) = − log2x, we flip the graph of log2xover the x-axis

f (x) = –log2 x

x

y

2 1

1 2 3

–2 –1

(.5, 1) (1, 0)

(2, –1)

EXAMPLE 13.12 Sketch the graph of g(x) = − log2(−x) What is the domain of g?

SOLUTION We can take the log of positive numbers only, so the domain of this function is x < 0 The

graph of g can be obtained by reflecting the graph of f from the previous example over the y-axis

x

y

2 1

1 2

–2

–2 –1 –1

(–.5, 1) (–1, 0)

(–2, –1)

EXAMPLE 13.13 Sketch the graph of h(x) = log327x2

SOLUTION What is the domain of h? We can take the log of positive numbers only, so the domain of h

is all nonzero x Rewrite h(x) so that it looks more familiar

h(x) = log327x2= log327 + log3x2= 3 + log3x2 h(x)is an even function (because h(−x) = h(x)), so its graph is symmetric about the y-axis Therefore, if we graph y = 3 + 2 log3xand reflect this graph about the y-axis, we’ll

be all set

How do we graph y = 3 + 2 log3x? We’ll start with the graph of log3xand build it

up from there log3xand 3xare inverse functions, so their graphs are reflections about the line y = x

–1 –1 1

1

x

y

y = x

y = log3 x

y = 3x

Figure 13.8

Multiplying log3x by 2 stretches the graph vertically, and adding 3 shifts the graph up 3 units The graph of y = 3 + 2 log3xis shown below

y

y = 3 + 2 log3 x

Figure 13.9

Should we want the exact value of the x-intercept, we can get it by solving the equation

0 = 3 + 2 log3x We obtain

log3x =−3

3log3 x

= 3−3/2

3 The graph of h(x) is given on the following page

y

h (x) = log3 (27x2 )

–1 3√3 1 3√3

P R O B L E M S F O R S E C T I O N 1 3 4

In Problems 1 through 5, sketch the graph of the function without the use of a computer

or graphing calculator.

1 y = ln(x + 1)

2 y = ln(x2)

3 y = | ln x|

4 y = ln |x|

5 y = ln(x1)

6 Sketch a rough graph of y = ln x − ln(x3) + 4 ln(x2) (Hint: This will be straight-forward after you have rewritten in the form y = K ln x, where K is a constant.)

Differentiating Logarithmic and Exponential Functions

Why is ln x the most frequently used logarithm in calculus? Why is it known as the “natural” logarithm? What’s so natural about it? You’ll begin to understand as we investigate its derivative

467

Exploratory Problem for Chapter 14

The Derivative of the Natural Logarithm

In this exploratory problem, you’ll look at the derivative of ln x both qualitatively and numerically

Graphical Analysis:Sketch the graph of f (x) = ln x and below that sketch its derivative

Numerical Analysis:We know that f(b) = limh→0f (b+h)−f (b)h ,

so f(b) ≈f (b+h)−f (b)h for h very small In a collaborative effort with a handful of your most trusted colleagues, calculate numer-ical approximations of f(b)for various b’s by choosing a small

hand calculating the slope of relevant secant lines Complete the accompanying table using the results

x f(x)(approximated) 0.1

0.5 1 2 3 4 5 6

Conjecture:The derivative of ln x is

The Derivative of ln x

In the exploratory problem you probably conjectured that the derivative of ln x is1x This is only a conjecture; the numerical and graphical evidence is strong, and the result is beautiful

enough to have a pull of its own, but we cannot prove the conjecture using numerical

approximations and calculators Mathematicians build on solid ground We give a strong argument below

Graphical Analysis

x

f (x) = lnx f ′ (x) = lnx

f is always increasing, so f ′ > 0 everywhere.

f is always concave down, so f ′ is decreasing.

(i) y = ln x (ii) y = ln x d

dx

d dx

Figure 14.1

Solid Argument

We will show that dxd ln x =1x by using the inverse relation between ln x and exand the fact that d

dxex= ex (Because e was defined so that d

dxex= exand ln x was defined to be

the inverse of ex, we are building our argument on a solid foundation.)

Below are graphs of exand ln x with an arbitrary point (b, ln b) labeled on the graph

of ln x For any point (b, c) on the graph of ln x there is the point (c, b) on the graph of ex

because these functions are inverses

y

y = ln x

y = e x

(c, b)

(b, c) = (b, ln b)

Figure 14.2

We are looking for the slope of the tangent line to ln x at x = b Let L1be the tangent line to exat (c, b) and L2be the tangent line to ln x at (b, c).1

y

L1

L1

L2

L2

y = ln x

y = e x

(c, b)

(b, c)

Figure 14.3

Becausedxdex= ex(i.e., the slope at any point on the graph of exis given by the y-coordinate

of that point), the slope of the tangent line to exat (c, b) is b Since the roles of x and y are interchanged on the graph of ln x, the slope of the tangent line at (b, c) is1b, as conjectured Let’s make this argument more explicit We assert that the slopes of L1and L2are reciprocals using the following reasoning If (b, c) and (s, t) are two points on L2, then (c, b) and (t, s) are two points on L1 The slope of L2is t −c

s−b; the slope of L1iss−b

t −c

the slope of L1

Conclusion:

d

x This result deserves some celebration! It’s nice, neat, clean It’s very beautiful.2

EXERCISE 14.1 Differentiate f (x) = 3 ln 2x (Hint: Pull apart the expression on the right so it is written

as a sum The two terms of the sum will be simple to differentiate.)

EXERCISE 14.2 Differentiate f (x) = log x by converting log x to log base e

The Derivative of logbx

Let’s find the derivative of logbx We’ll do this by converting logbxto an expression using natural logarithms and then use what we’ve just learned about the derivative of ln x

1 If an invertible function is differentiable at the point (c, b) with nonzero derivative, its inverse function will be differentiable

at the point (b, c) This should make sense from a graphical point of view.

2 There is even more to celebrate Recall that we have shown that d

dx x n = nx n−1 for n = −1, 0, 1, 2, It is also true that d

dx x n = nx n−1 for any constant n We have not yet shown this, but we will Using this fact, we can find a function whose derivative

is x k for any constant k, except k = −1 d

dx

 x k+1

k+1



=k+1x k = x k for k = −1 Now we can also find a function whose derivative

−1