MATH 180 FALL 1995

ASSIGNMENT #5

1.7, #2. 3 ln e + ln(1/e) = 3 ln e − ln e = 3 − 1 = 2.

1.7, #11. 2e^{3x} = 4e^{5x} is equivalent to \frac{1}{2} = e^{(5−3)x} = e^{2x}. Thus x = \frac{1}{2} ln \frac{1}{2} = −.34657.

1.7, #15. P = 10e^{0.917t} = 10(2.502)^{t} represents exponential growth since the base 2.502 > 1, or equivalently, since 0.917 > 0.

1.7, #17. P = 79e^{−2.5t} = 79(.0821)^{t} represents exponential decay since the base .0821 < 1, or equivalently, since −2.5 < 0.

1.7, #27. This is similar to 1.6, #9 in Assignment #4. Apply the natural logarithm to both sides of P = P_{0}e^{kt} and solve for t:

t = \frac{1}{k} (ln P − ln P_{0}).

Comments: The problems in this assignment are straightforward exercises in exponential and logarithm functions. Review the material if your writeup is not perfect.