James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 1.4 Exponential Functions 51
Solution
(a) The mass is initially 24 mg and is halved during each 25-year period, so
ms0d − 24
ms25d − 1 2 s24d
ms50d − 1 2 ? 1 2 s24d − 1 2 2 s24d
ms75d − 1 2 ? 1 2 2 s24d − 1 2 3 s24d
ms100d − 1 2 ? 1 2 3 s24d − 1 2 4 s24d
From this pattern, it appears that the mass remaining after t years is
30
mstd − 1
2 ty25 s24d − 24 ? 22ty25 − 24 ? s2 21y25 d t
This is an exponential function with base b − 2 21y25 − 1y2 1y25 .
m=24 · 2 _t/25 m=5
(b) The mass that remains after 40 years is
ms40d − 24 ? 2 240y25 < 7.9 mg
0 100
FIGURE 12
(c) We use a graphing calculator or computer to graph the function mstd − 24 ? 2 2ty25
in Figure 12. We also graph the line m − 5 and use the cursor to estimate that mstd − 5
when t < 57. So the mass of the sample will be reduced to 5 mg after about 57 years. ■
The Number e
Of all possible bases for an exponential function, there is one that is most convenient
for the purposes of calculus. The choice of a base b is influenced by the way the graph
of y − b x crosses the y-axis. Figures 13 and 14 show the tangent lines to the graphs of
y − 2 x and y − 3 x at the point s0, 1d. (Tangent lines will be defined precisely in Section
2.7. For present purposes, you can think of the tangent line to an exponential graph at a
point as the line that touches the graph only at that point.) If we measure the slopes of
these tangent lines at s0, 1d, we find that m < 0.7 for y − 2 x and m < 1.1 for y − 3 x .
y
y=2®
y y=3®
mÅ0.7
mÅ1.1
1
1
0
x
0
x
figure 13 figure 14
It turns out, as we will see in Chapter 3, that some of the formulas of calculus will be
greatly simplified if we choose the base b so that the slope of the tangent line to y − b x
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