A radioactive substance decays so that after t years, the amount remaining, expressed as a percent of the original amount, is A(t)= 100(1. 2)^(-t).
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a) What is the half-life for this substance? Round to 2 decimal places.
b) Determine the rate of decay after 5 years. Round to 2 decimal places.
Many thanks!!
a) you want to know when the amount left is 50
50 = 100(1.2)^-t
.5 = 1.2^(-t)
log both sides
log .5 = -t log 1.2
-t = log1.2/log.5 = -.263
t = appr .263 years or appr 3.2 months
b) A'(t) = -100ln1.2(1.2)^-t
replace t with 5 and evaluate
0.5 = 1.2^-t
log 0.5 = -t log 1.2
-0.30 = -t * 0.0792
t =3.79 years
A = 1.2^-t
dA/dt = -t *1.2^(-t-1)
at t = 5
dA/dt =-5 * 1.2^(-6) = -5 * 0.335 = -1.67
a) To find the half-life for this substance, we need to determine the time it takes for the amount remaining to be half of the original amount.
The original amount, expressed as a percent, is 100%. So, we need to find the value of t when A(t) is equal to 50%.
Setting A(t) equal to 50% and solving for t:
50 = 100(1.2)^(-t)
Dividing both sides by 100:
0.5 = (1.2)^(-t)
Taking the logarithm of both sides:
log(0.5) = log((1.2)^(-t))
Using the logarithmic property log(a^b) = b*log(a):
log(0.5) = -t*log(1.2)
Dividing both sides by the logarithm of 1.2:
t = log(0.5) / log(1.2)
Using a scientific calculator or an online calculator to evaluate the logarithms:
t ≈ 6.91
Therefore, the half-life for this substance is approximately 6.91 years.
b) To determine the rate of decay after 5 years, we need to evaluate the derivative of A(t) with respect to t.
A(t) = 100(1.2)^(-t)
Taking the derivative of A(t) with respect to t:
A'(t) = -100*log(1.2)*(1.2)^(-t)
Substituting t = 5 into the derivative:
A'(5) = -100*log(1.2)*(1.2)^(-5)
Using a scientific calculator or an online calculator to evaluate the logarithm and exponents:
A'(5) ≈ -11.82
Therefore, the rate of decay after 5 years is approximately -11.82% per year.