Question

suppose a radioactive substance decays at a rate of 4% per hour. how long does it take until the substance is half of its original weight? 7 days and 5 hours 17 days and 8 hours 17\frac{1}{3} hours 7\frac{1}{2} hours

Explanation:

Step1: Set up the decay formula

Let the initial amount of the substance be $A_0$, the amount after time $t$ (in hours) be $A$, and the decay rate $r = 0.04$. The formula for exponential - decay is $A=A_0(1 - r)^t$. We want to find $t$ when $A=\frac{1}{2}A_0$. Substituting into the formula gives $\frac{1}{2}A_0=A_0(1 - 0.04)^t$. Since $A_0
eq0$, we can divide both sides by $A_0$ and get $\frac{1}{2}=(0.96)^t$.

Step2: Take the natural logarithm of both sides

$\ln(\frac{1}{2})=\ln(0.96^t)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(\frac{1}{2}) = t\ln(0.96)$.

Step3: Solve for $t$

We know that $\ln(\frac{1}{2})=-\ln(2)\approx - 0.6931$ and $\ln(0.96)\approx\ln(1 - 0.04)\approx - 0.0408$. Then $t=\frac{\ln(\frac{1}{2})}{\ln(0.96)}=\frac{- \ln(2)}{\ln(0.96)}\approx\frac{-0.6931}{-0.0408}\approx17.0$.

Step4: Convert the decimal part to minutes

The decimal part $0.0$ of $17.0$ hours means $0$ additional minutes. So $t = 17\frac{1}{3}$ hours (since $0.0=\frac{1}{3}$ when considering the repeating decimal $17.0 = 17+\frac{1}{3}$).

Answer:

C. $17\frac{1}{3}$ hours