modeling with mathematics one hundred grams of radium are stored in a container. the amount r (in grams) of…

modeling with mathematics one hundred grams of radium are stored in a container. the amount r (in grams) of radium present after t years is modeled by r = 100e^(-0.00043t). after how many years will only 5 grams of radium be present? round your answer to the nearest whole year. it will take about years for only 5 grams of radium to be present.

modeling with mathematics one hundred grams of radium are stored in a container. the amount r (in grams) of radium present after t years is modeled by r = 100e^(-0.00043t). after how many years will only 5 grams of radium be present? round your answer to the nearest whole year. it will take about years for only 5 grams of radium to be present.

Answer

Explanation:

Step1: Substitute values into formula

We are given $R = 5$ and the formula $R = 100e^{-0.00043t}$. So, $5=100e^{-0.00043t}$.

Step2: Isolate the exponential term

Divide both sides by 100: $\frac{5}{100}=e^{-0.00043t}$, which simplifies to $0.05 = e^{-0.00043t}$.

Step3: Take natural - logarithm of both sides

$\ln(0.05)=\ln(e^{-0.00043t})$. Since $\ln(e^x)=x$, we have $\ln(0.05)=- 0.00043t$.

Step4: Solve for t

$t=\frac{\ln(0.05)}{-0.00043}$. We know that $\ln(0.05)\approx - 2.995732$. Then $t=\frac{-2.995732}{-0.00043}\approx6967$.

Answer:

6967