Math 1100-006
§5.3
1. (2 points) (Problem 34) Solve the following exponential equation:
2000(1 − 1.005−x )
67, 405 =
0.005
Solution:
(67, 405)(0.005) = 2000(1 − 1.005−x )
67, 405(0.005)
= 1 − 1.005−x
2000
0.1685125 = 1 − 1.005−x
0.1685125 − 1 = −1.005x
−0.8314875 = −1.005−x
0.8314875 = 1.005−x
log(0.8314875) = log(1.005−x )
log(0.8314875) = −x log(1.005)
log(0.8314875)
−x =
log(1.005)
−0.080144
x=−
0.0021661
x = 36.9992
2. (2 points) (Problem 53) Solve the logarithmic equation:
ln x + ln 3x = 3
Solution:
ln x + ln 3x = 3
ln(3x2 ) = 3
e3 = 3x2
e3
= x2
3
6.695 = x2
x = ±2.5875
Solution Key
3. (Problem 105) By using data from the U.S. Bureau of Labor Statistics for the years
1968-2002, the purchasing power P of a 1983 dollar can be modeled with the function
P (t) = 3.818e−0.0506t
where t is the number of years past 1960.
(a) (4 points) Find P(10) and P(70) and for each write a sentence that interprets its
meaning.
Solution:
P (10) = 3.818e−0.0506(10)
= 3.818e−0.506
= 3.818(0.6029)
= 2.302
P (10) = 2.30 means that the purchasing power of a 1983 dollar in 1970 was
approximately $2.30. So one dollar had less purchasing power in 1983 than it
did in 1970, 13 years earlier.
P (70) = 3.818e−0.0506(70)
= 3.818e−3.542
= 3.818(0.02896)
= 0.1105
P (70) = 0.11 means that the purchasing power of a 1983 dollar will be approximately $0.11 in 2030.
(b) (2 points) How long before the purchasing power of a 1983 dollar is $0.25?
Solution: We need to solve 0.25 = P (t) for t:
0.25 = 3.818e−0.0506(t)
0.25
= e−0.0506(t)
3.818
0.06548 = e−0.0506(t)
ln(0.06548) = −0.0506t
−2.726
=t
−0.0506
t = 53.874
53 years after 1960 is 2013. So the purchasing power of a 1983 dollar will reach
$0.25 in late 2013.
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