Calculus: Analytic Geometry and Calculus, with Vectors

9.1 Exponentials and logarithms 487

the graph of y = loge x at the point (1,0). Remark: This is not a dull problem,
because moderately careful work produces very good results. To check one of
the results, let f (x) = 2' so f (x) = ex l°g 2 so f' (x) = e° log 2(log 2) and f'(0) _
log 2 = 0.693.
7 Apply the procedure of Problem 6 to the equations y = 3' and y = logs x.
Try to find a way to use a table or slide rule to check the numerical results.
8 Many persons with scientific interests have (and should have) log-log
slide rules and should know or learn how to set them to obtain values of as when
2 < a < 3 and -1 < x <= 1. These persons can quickly produce a good graph
of y = (2.5)° over the interval -1 5 x < 1 and use it to obtain a good estimate
of the slope at the point (0,1). Repeating the process for y = (2.7S)° and using
ideas involving interpolation lead to quite good approximations to a base e for
which the graph of y = e° has slope 1 at the point (0,1). This base e is the famous

9 Tell why

fo1

2z dx and fo1 3' dx exist. Then sketch appropriate graphs

and use them to obtain rough estimates of the values of these integrals. Finally,

check the estimates with the aid of the formula

I^1 ax dx = f e' log a dx =^11 e' log a 1 l

0 o log a Jo

1 (e loga - 1) =a- 1

log a 1) log a

Tell why foI 100° dx could not be 0.05 and could not be 500.

Discover reasons why we could suspect that

J

a'dt<1

2a°

when a > 1 and x > 0.

12 Sketch graphs of the equations

(a) y = 2-1/z% (b) y

Remark: More detailed information about the graph of the equation y = il/='

will appear later.

13 In Problem 4 of Section 12.6, we shall discover that if z > 0, then

(1) log z! = log 2a + (z + 'ff') log z - z + E(z),

where E(z) is a number for which

1 1 1 1 1

(2) 12z 36071 E(z) <12z 360z3 1260z6

Show that (1) can be put in the form

(3)

and hence that

(4)

log z! =log 2za+logz'-z-I-E(z)

z! = 2zlr z°e-'ez(s).