- #1
karmatic
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Homework Statement
The amount (A) of cesium-137 remaining after t years is given by
A=A[itex]_{0}[/itex]*2[itex]^{\frac{-t}{30.3}}[/itex]
where A[itex]_{0}[/itex] is the initial amount. In what year will the cesium-137 be 10% of that which was released at the Chernobyl disaster in 1986?
Homework Equations
x[itex]^{\frac{m}{n}}[/itex]=(x[itex]^{\frac{1}{n}}[/itex])[itex]^{m}[/itex]
The Attempt at a Solution
plug in my 0.1 on the LHS
0.1A[itex]_{0}[/itex]=A[itex]_{0}[/itex]*2[itex]^{\frac{-t}{30.3}}[/itex]
A[itex]_{0}[/itex] on both sides cancels out
0.1=2[itex]^{\frac{-t}{30.3}}[/itex]
convert decimal to a fraction
[itex]\frac{1}{10}[/itex]=2[itex]^{\frac{-t}{30.3}}[/itex]
take the reciprocal of the RHS to convert the negative power to a positive
[itex]\frac{1}{10}[/itex]=[itex]\frac{1}{2^{\frac{t}{30.3}}}[/itex]
take the reciprocal of both sides to remove the fractions
10=2[itex]^{\frac{t}{30.3}}[/itex]
isolate the variable t and calculate the remaining values
10=(2[itex]^{\frac{1}{30.3}}[/itex])[itex]^{t}[/itex]
10=1.02313981[itex]^{t}[/itex]
then this part I'm unsure if I'm doing right, and I'm not aware of how I should be trying to check my work
t=[itex]\frac{log 10}{log 1.02313981}[/itex]
t=100.6544213
1986+100=2086?
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