SOLUTION: Please help. These log word problems are the worst.
A certain radioactive isotope decays at a rate of .25% annually. Determine the half-life of this isotope, to the nearest year
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A certain radioactive isotope decays at a rate of .25% annually. Determine the half-life of this isotope, to the nearest year
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Question 151233: Please help. These log word problems are the worst.
A certain radioactive isotope decays at a rate of .25% annually. Determine the half-life of this isotope, to the nearest year. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A certain radioactive isotope decays at a rate of .25% annually. Determine the half-life of this isotope, to the nearest year.
:
The half-life formula: A = Ao*2^(-t/h)
where:
Ao - initial amt
A = resulting amt
t = time
h = half life of the substance
;
In this problem:
A = .75 (amt remaining after a 25% loss)
Ao = 1
t = 1
h = half life
:
We can write it:
1* 2(-1/h) = .75
:
ln(2^(-1/h)) = ln(.75); Find the nat log of both sides
: *ln(2) = ln(.75); the log equiv of exponents
.693147* = -.287682 = -.287682
-.693147 = -.287682h; multiplied both sides by h
:
h =
h = +2.4 years is the half-life of this isotope
h = 2 yrs
:
:
Check solution on a calc: enter 2^(-1/2.4) = .749 ~ .75 (left after 25% loss)
