SOLUTION: The half-life of cobalt - 60 is 5.27 years. Starting with a sample of 150 mg, after how many years is 20 mg left?

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Question 1176085: The half-life of cobalt - 60 is 5.27 years. Starting with a sample of 150 mg, after how many years is 20 mg left?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the half life of cobalt-60 is 5.27 years.

to find the annual rate of decay, use the exponential formula to get:

.5 = x ^ 5.27

take the 5.27th root of both sides of this equation to get:

.5 ^ (1/5.27) = x

solve for x to get:

x = .8767556206

to confirm that value of x is good, replace x in the original equation and solve for y.

you get y = .8767556206 ^ 5.27 = .5

this confirms the value of x is good.

you can now use that value of x to solve the problem.

if y = 20 when x = 150, then the formula becomes:

20 = 150 * .8767556206 ^ n

divide both sides of that equation by 150 to get:

20/150 = .8767556206 ^ n

take the log of both sides of that equation to get:

log(2/15) = log(.8767556206 ^ n)

since log(x ^ a) = a * log(x), that equation becomes:

log(2/15) = n * log(.8767556206)

solve for n to get:

n = log(2/15) / log(.8767556206) = 15.31931344.

confirm by replacing n in the original equation by that value and solving for y.

you get:

y = 150 * .8767556206 ^ 15.3191344 = 20.

that confirms the value of n is good.

15.3191344 years is your solution.



Answer by ikleyn(52799) About Me  (Show Source):
You can put this solution on YOUR website!
.
The half-life of cobalt - 60 is 5.27 years. Starting with a sample of 150 mg, after how many years is 20 mg left?
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Since the half-line is given in the problem, you can write the decay formula in this form

    M = M%5B0%5D%2A%281%2F2%29%5E%28t%2F5.27%29.


In this equation,  M%5B0%5D  is the starting mass of the radioactive material;  

M is the current mass after t yeras of decay.



In the problem, you are given  M%5B0%5D = 150 mg  and  M = 20 mg, and they want you find t.


So, your equation is

    20 = 150%2A%281%2F2%29%5E%28t%2F5.27%29.


Divide both sides by 150

    20%2F150 = %281%2F2%29%5E%28t%2F5.27%29,   or   0.1333 = %281%2F2%29%5E%28t%2F5.27%29.


Take logarithm base 10 of both sides

    log(0.1333) = %28t%2F5.27%29%2Alog%28%280.5%29%29


and express t from the last equation

    t = %285.27%2Alog%28%280.1333%29%29%29%2Flog%28%280.5%29%29.


Now use your calculator

    t = 15.32 years.     ANSWER

Solved.

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To learn more on the subject,  look into the lesson
    - Radioactive decay problems
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Logarithms".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.