SOLUTION: A satellite has a radioactive isotope power supply. The power output in watts is given by the equation P = 50e^(t/250), where P is the power in watts and t is the time in days.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: A satellite has a radioactive isotope power supply. The power output in watts is given by the equation P = 50e^(t/250), where P is the power in watts and t is the time in days.       Log On


   

Question 1032718: A satellite has a radioactive isotope power supply. The power output in watts is given by the equation P = 50e^(t/250), where P is the power in watts and t is the time in days.
a ) What is the half life of the power supply?
b) The equipment aboard the satellite requires 10 watts of power to operate properly. What is the operational life of the satellite?


I am having trouble setting up these equations properly.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Question:
A satellite has a radioactive isotope power supply. The power output in watts is given by the equation P = 50e^(t/250), where P is the power in watts and t is the time in days.
a ) What is the half life of the power supply?
b) The equipment aboard the satellite requires 10 watts of power to operate properly. What is the operational life of the satellite?

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Since this is decaying, the exponent needs to have a negative in it. So it should be
P = 50e^(-t/250)


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Part (a) What is the half life of the power supply?

Plug in t = 0 to get...
P = 50e^(-t/250)
P = 50e^(0/250)
P = 50e^(0)
P = 50*1
P = 50
So at t = 0 days (aka the beginning), the power output in watts is 50 watts

So we start off with 50 watts. The question is: how long will it take to get to 25 watts? Which is half of the original output.

Let's find out. Plug in P = 25 and solve for t.

P = 50e^(-t/250)
25 = 50e^(-t/250)
25/50 = [50e^(-t/250)]/50
0.5 = e^(-t/250)
e^(-t/250) = 0.5
Ln[e^(-t/250)] = Ln[0.5]
(-t/250)*Ln[e] = Ln[0.5]
(-t/250)*1 = Ln[0.5]
-t/250 = Ln[0.5]
t = -250*Ln[0.5]
t = 173.286795

It takes approximately 173.286795 days for the amount of 50 watts to turn into 25 watts.
So the half-life is approximately 173.286795 days

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Part (b) The equipment aboard the satellite requires 10 watts of power to operate properly. What is the operational life of the satellite?

Now we need to find out how long this power supply will work as long as P >= 10. Let's plug in P = 10, then solve for t, to find out

P = 50e^(-t/250)
10 = 50e^(-t/250)
10/50 = [50e^(-t/250)]/50
0.2 = e^(-t/250)
e^(-t/250) = 0.2
Ln[e^(-t/250)] = Ln[0.2]
(-t/250)*Ln[e] = Ln[0.2]
(-t/250)*1 = Ln[0.2]
-t/250 = Ln[0.2]
t = -250*Ln[0.2]
t = 402.359478

So the power supply will last for approximately 402.359478 days til it reaches 10 watts. After that, it won't operate properly since it won't produce enough watts.

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Summary:
Answer to part (a): 173.286795 days (this is approximate)
Answer to part (b): 402.359478 days (this is approximate)