SOLUTION: I worked this problem out but would like to check with you if I'm vaguely correct:
The half-life of radium-226 is 1600 years. Suppose we have a 22-mg sample.
a. Find a function
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Question 48240This question is from textbook College Algebra
: I worked this problem out but would like to check with you if I'm vaguely correct:
The half-life of radium-226 is 1600 years. Suppose we have a 22-mg sample.
a. Find a function that models the mass remaining after t years
Using the model for radioactive decay mt = m0e^-rt:
m0 = 22mg, r=(ln2/1600) = -0.00043
m(t) = 22e^-0.00043t
b. How much of the sample will remain after 4000 years?
t = 4000
m(4000) = 22e^(-0.00043)(4000) = 3.94 mg
c. After how long will only 18 mg of the sample remain?
m(t) = 18
22e^-0.00043t = 18
e^-0.00043t = 9/11
Ln e^-0.00043t = ln 9/11
0.00043t = ln 9/11
t = -(ln(9/11)/0.0043) = 46.7 years
Thanks a million(factored many times) for your help!
This question is from textbook College Algebra
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The half-life of radium-226 is 1600 years. Suppose we have a 22-mg sample.
a. Find a function that models the mass remaining after t years
Using the model for radioactive decay mt = m0e^-rt:
m0 = 22mg, r=(ln2/1600) = -0.00043
-----------------------------------
A rate cannot be negative.
You probably mean r=0.00043
----------------------------
m(t) = 22e^-0.00043t
b. How much of the sample will remain after 4000 years?
t = 4000
m(4000) = 22e^(-0.00043)(4000) = 3.94 mg
c. After how long will only 18 mg of the sample remain?
m(t) = 18
22e^-0.00043t = 18
e^-0.00043t = 9/11
Ln e^-0.00043t = ln 9/11
0.00043t = ln 9/11
-----------------------
I think you need the following:
ln e^-0.00043t = ln 9/11
-0.00043t = ln(9/11)
t=[ln(9/11)]/(-0.00043)
t=466.676 years
Cheers,
Stan H.
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