The rate of decay of radium is proportional to
the amount present at any time.
(In other words, dq/dt = kq.).
If after ten years a quantity of radium has
decreased to 4.979 grams, and at the end of
an addditional ten years it has decreased to
4.958 grams, how many grams were
originally present?
Do not round the value of k , but round your
final answer to the nearest thousandth.

Question

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Given that the rate of decay of radium is proportional to the amount present at any time, we can express this relationship as:

$$ \frac{dq}{dt} = kq $$

where:
- $$ q $$ is the amount of radium present at time $$ t $$,
- $$ k $$ is the decay constant.

We are given that after ten years, the quantity of radium has decreased to 4.979 grams, and after an additional ten years, it has decreased to 4.958 grams.

Let's denote the original amount of radium as $$ q_0 $$ and the decay constant as $$ k $$.

From the given information, we can set up the following equations:

1. After ten years:
$$ q_0 \cdot e^{-10k} = 4.979 $$

2. After twenty years:
$$ q_0 \cdot e^{-20k} = 4.958 $$

We can solve these equations to find the original amount of radium $$ q_0 $$ and the decay constant $$ k $$.

Let's solve the system of equations to find the values of $$ q_0 $$ and $$ k $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4.979=q_{0}e^{-10k}\4.958=q_{0}e^{-20k}\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}q_{0}=\frac{4979e^{10k}}{1000}\4.958=q_{0}e^{-20k}\end{array}\right.$$
- step2: Substitute the value of $$q_{0}:$$
  $$4.958=\frac{4979e^{10k}}{1000}e^{-20k}$$
- step3: Simplify:
  $$4.958=\frac{4979}{1000e^{10k}}$$
- step4: Swap the sides:
  $$\frac{4979}{1000e^{10k}}=4.958$$
- step5: Convert the expressions:
  $$\frac{4979}{1000e^{10k}}=\frac{2479}{500}$$
- step6: Rewrite the expression:
  $$1000e^{10k}=\frac{4979\times 500}{2479}$$
- step7: Divide the terms:
  $$1000e^{10k}=\frac{2489500}{2479}$$
- step8: Multiply by the reciprocal:
  $$1000e^{10k}\times \frac{1}{1000}=\frac{2489500}{2479}\times \frac{1}{1000}$$
- step9: Multiply:
  $$e^{10k}=\frac{4979}{4958}$$
- step10: Take the logarithm of both sides:
  $$\ln{\left(e^{10k}\right)}=\ln{\left(\frac{4979}{4958}\right)}$$
- step11: Evaluate the logarithm:
  $$10k=\ln{\left(\frac{4979}{4958}\right)}$$
- step12: Divide both sides:
  $$\frac{10k}{10}=\frac{\ln{\left(\frac{4979}{4958}\right)}}{10}$$
- step13: Divide the numbers:
  $$k=\frac{\ln{\left(\frac{4979}{4958}\right)}}{10}$$
- step14: Simplify:
  $$k=\frac{\ln{\left(4979\right)}-\ln{\left(4958\right)}}{10}$$
- step15: Substitute the value of $$k:$$
  $$q_{0}=\frac{4979e^{10\times \frac{\ln{\left(4979\right)}-\ln{\left(4958\right)}}{10}}}{1000}$$
- step16: Substitute back:
  $$q_{0}=\frac{4979\left(e^{\frac{\ln{\left(4979\right)}-\ln{\left(4958\right)}}{10}}\right)^{10}}{1000}$$
- step17: Calculate:
  $$q_{0}\approx 5.000089$$
- step18: Calculate:
  $$\left\{ \begin{array}{l}k=\frac{\ln{\left(4979\right)}-\ln{\left(4958\right)}}{10}\q_{0}\approx 5.000089\end{array}\right.$$
- step19: Check the solution:
  $$\left\{ \begin{array}{l}k=\frac{\ln{\left(4979\right)}-\ln{\left(4958\right)}}{10}\q_{0}\approx 5.000089\end{array}\right.$$
- step20: Rewrite:
  $$\left(k,q_{0}\right)\approx \left(\frac{\ln{\left(4979\right)}-\ln{\left(4958\right)}}{10},5.000089\right)$$
The solution to the system of equations is:

$$ k \approx 0.000423 $$
$$ q_0 \approx 5.000089 $$

Therefore, the original amount of radium was approximately 5.000089 grams. Rounding to the nearest thousandth, the original amount of radium was approximately 5.000 grams.
The original amount of radium was approximately 5.000 grams.

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