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 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. 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 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. it will take about years for only 5 grams of radium to be present.

Answer

Explanation:

Step1: Set up the equation

Set $R = 5$ in the formula $R = 100e^{-0.00043t}$, so we get $5=100e^{-0.00043t}$.

Step2: Simplify the equation

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

Step3: Take the natural - logarithm of both sides

$\ln(0.05)=\ln(e^{-0.00043t})$. Since $\ln(e^{x}) = x$, the right - hand side simplifies to $- 0.00043t$. So, $\ln(0.05)=-0.00043t$.

Step4: Solve for $t$

We know that $\ln(0.05)\approx - 2.995732$. Then $t=\frac{\ln(0.05)}{-0.00043}$. Substitute the value of $\ln(0.05)$ into the formula: $t=\frac{-2.995732}{-0.00043}\approx6967$.

Answer:

$6967$