EXPONENTIAL AND LOGARITHMIC FUNCTIONS

PROBLEM 6.

A satellite has a radioisotope power supply. The power output in watts is given by the equation

P = 50e^(-t/250)

where t is the time in days and e is the base of natural logarithms.

a. How much power will be available at the end of one year?

Solution: Applying the given equation, we have

P = 50e^(-365/250)

= 5Oe^(-1.46)

= 50 x 0.232

= 11 .6

Thus approximately 11.6 watts will be available at the end of one year.

b. What is the half-life of the power supply? In other words, how long will it take for the power to drop to half its original strength?

Solution:
To find the half-life, we solve the equation

25 = 50e^(-t/250)

for t and obtain

- t/250 = ln O.5

= -0.693

t = 250 x 0.693

= 173.

Thus the half-life of the power supply is approximately 173 days. (Note that In x is a shorter expression for log[base e] x.)

c. The equipment aboard the satellite requires 10 watts of power to operate properly. What is the operational life of the satellite?

Solution: Solving the equation

10 = 50e^(-t/250)

for t gives

-t/250 = ln(10/50)

= In 0.2

= - 1 .609

t = 250 x 1.609

= 402.

Hence the operational life of the satellite is 402 days.

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