Calculus HW help!?

Method 1 :

ln(A/Aₒ) = -kt

After 10 years:
ln(0.9) = -k(10) …… [1]

After a period of half-life:
ln(0.5) = -k(t½) …… [2]

[2] / [1]:
t½ / 10 = ln(0.5 / ln(0.9)
t½ = 10 [ln(0.5)/ln(0.9)]
Half-life, t½ = 66 years

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Method 2:

Let n be the number of periods of half-life.

0.9 = (1/2)ⁿ
log(0.9) = log[(1/2)ⁿ]
log(0.9) = n log(0.5)
n = log(0.9) / log(0.5)
1 / n = log(0.5) / log(0.9)

Half-life = 10 / n = 10 [log(0.5) / log(0.9)] = 66 years

Let I0 be the unit at the exploion in 1970, it was 0.9 I0 in 1980 after 10 yearsI. Use I=i0 exp (-at) Put the values I/I0 = 9/10= exp(-a 10) 10 years.
Calculate a value. -10 a= ln 0.9=-0.01. a= 0.01. Now we can find the half life.
I/I0=1/2 = exp (0.01 t). ln 1./2=-0.69=0.01 t.
t=69 years. I hope that it is correct,

R(t) = R(0) x (0.9)^(t/10)

We require when R(t)/R(0) = 0.5

so, when (0.9)^(t/10) = 0.5

=> (t/10)ln(0.9) = ln(0.5)

i.e. t = 10ln(0.5)/ln(0.9) => 65.8 years

Hence, roughly September, 2035

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