Question

math1111: college algebra (11446)
lesson 28 - exponential and logarithmic equations with applic
lesson 28 - exponential and logarithmic equations with applications
score: 88/100 answered: 18/20
question 20
the half-life of radium-226 is 1590 years. use the the formula $a(t) = a_0e^{kt}$ to model the situation.
(a) find the exact value of $k$, the rate of decay per year. round to 6 decimal places or enter an exact answer.
$k = $
(b) if a sample contains 200 mg, how many mg will remain after 3000 years? round to 2 decimal places.
amount: mg
(c) if a sample contains 200 mg, how long will it take to decay to 187.5 mg? round to the nearest year.
time: years
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Explanation:

Step1: Find k (Part a)

Half - life means when \(t = 1590\), \(A(t)=\frac{A_0}{2}\). Substitute into \(A(t)=A_0e^{kt}\):
\(\frac{A_0}{2}=A_0e^{k\times1590}\)
Divide both sides by \(A_0\) (assuming \(A_0
eq0\)): \(\frac{1}{2}=e^{1590k}\)
Take natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{1590k})\)
Using \(\ln(e^x)=x\), we get \(\ln(\frac{1}{2}) = 1590k\)
So \(k=\frac{\ln(\frac{1}{2})}{1590}=\frac{-\ln(2)}{1590}\approx\frac{- 0.693147}{1590}\approx - 0.000436\)

Step2: Find amount after 3000 years (Part b)

We know \(A_0 = 200\), \(t = 3000\), and \(k=\frac{-\ln(2)}{1590}\). Use \(A(t)=A_0e^{kt}\)
\(A(3000)=200\times e^{\frac{-\ln(2)}{1590}\times3000}\)
First, calculate the exponent: \(\frac{-\ln(2)\times3000}{1590}=\frac{-3000\ln(2)}{1590}\approx\frac{-3000\times0.693147}{1590}\approx - 1.317\)
Then \(A(3000)=200\times e^{-1.317}\approx200\times0.268\approx53.60\) (rounded to 2 decimal places)

Step3: Find time to decay to 187.5 mg (Part c)

We have \(A_0 = 200\), \(A(t)=187.5\), \(k=\frac{-\ln(2)}{1590}\). Use \(A(t)=A_0e^{kt}\)
\(187.5 = 200\times e^{\frac{-\ln(2)}{1590}t}\)
Divide both sides by 200: \(\frac{187.5}{200}=e^{\frac{-\ln(2)}{1590}t}\)
Simplify \(\frac{187.5}{200}=\frac{15}{16}\)
Take natural logarithm: \(\ln(\frac{15}{16})=\frac{-\ln(2)}{1590}t\)
Solve for \(t\): \(t=\frac{\ln(\frac{15}{16})\times(- 1590)}{\ln(2)}\)
\(\ln(\frac{15}{16})\approx - 0.0645\), so \(t=\frac{(-0.0645)\times(-1590)}{0.693147}\approx\frac{102.555}{0.693147}\approx148\) (rounded to nearest year)

Answer:

(a) \(k\approx\boldsymbol{-0.000436}\)
(b) amount \(\approx\boldsymbol{53.60}\) mg
(c) time \(\approx\boldsymbol{148}\) years