24. Bacteria Growth A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h there are 10,000 bacteria. At the end of 5 h there are 40,000 bacteria. How many bacteria were present initially? 25. Radon- 222 The decay equation for radon- 222 gas is known to be $$ y=y_{0} e^{-0.18 t} $$, with $$ t $$ in days. About how long will it take the amount of radon in a sealed sample of air to decay to $$ 90 \% $$ of its original value? 26. Polonium- 210 The number of radioactive atoms remaining after $$ t $$ days in a sample of polonium- 210 that starts with $$ y_{0} $$ radioactive atoms is $$ y=y_{0} e^{-0.005 t} $$. (a) Find the element's half-life.

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**Problem 24**

Let the number of bacteria at time $$ t $$ (in hours) be given by  
$$
y = y_0 e^{kt},
$$  
where $$ y_0 $$ is the initial number of bacteria and $$ k $$ is the growth rate.

We are given:  
- At $$ t = 3 $$ hours, $$ y = 10\,000 $$. Therefore,  
  $$
  10\,000 = y_0 e^{3k}.
  $$
- At $$ t = 5 $$ hours, $$ y = 40\,000 $$. Therefore,  
  $$
  40\,000 = y_0 e^{5k}.
  $$

Divide the second equation by the first to eliminate $$ y_0 $$:
$$
\frac{40\,000}{10\,000} = \frac{y_0 e^{5k}}{y_0 e^{3k}} \quad \Longrightarrow \quad 4 = e^{5k-3k} = e^{2k}.
$$

Taking the natural logarithm of both sides:
$$
\ln 4 = 2k \quad \Longrightarrow \quad k = \frac{\ln 4}{2}.
$$

Now substitute $$ k $$ back into the equation $$ 10\,000 = y_0 e^{3k} $$ to solve for $$ y_0 $$:
$$
y_0 = \frac{10\,000}{e^{3k}} = \frac{10\,000}{e^{3\left(\frac{\ln 4}{2}\right)}} = \frac{10\,000}{e^{\frac{3\ln 4}{2}}}.
$$

Recognize that $$ e^{\frac{3\ln 4}{2}} = 4^{3/2} $$. Since
$$
4^{3/2} = \left(4^{1/2}\right)^3 = 2^3 = 8,
$$
we have:
$$
y_0 = \frac{10\,000}{8} = 1250.
$$

---

**Problem 25**

For radon-222, the decay equation is given by
$$
y = y_0 e^{-0.18t},
$$
with $$ t $$ in days. We want the time $$ t $$ when the amount decays to $$ 90\% $$ of its original value, i.e., when
$$
y = 0.90\,y_0.
$$

Substitute into the equation:
$$
0.90\,y_0 = y_0 e^{-0.18t}.
$$

Cancel $$ y_0 $$ (assuming $$ y_0 \neq 0 $$):
$$
0.90 = e^{-0.18t}.
$$

Take the natural logarithm of both sides:
$$
\ln(0.90) = -0.18t.
$$

Solve for $$ t $$:
$$
t = -\frac{\ln(0.90)}{0.18}.
$$

Using the approximate value $$ \ln(0.90) \approx -0.10536 $$:
$$
t \approx \frac{0.10536}{0.18} \approx 0.5853 \text{ days}.
$$

---

**Problem 26**

For polonium-210, the decay is given by
$$
y = y_0 e^{-0.005t}.
$$
The half-life $$ T $$ is the time when $$ y = \frac{y_0}{2} $$. Substitute into the decay equation:
$$
\frac{y_0}{2} = y_0 e^{-0.005T}.
$$

Cancel $$ y_0 $$:
$$
\frac{1}{2} = e^{-0.005T}.
$$

Take the natural logarithm of both sides:
$$
\ln\left(\frac{1}{2}\right) = -0.005T.
$$

Since $$ \ln\left(\frac{1}{2}\right) = -\ln 2 $$, we have:
$$
-\ln 2 = -0.005T.
$$

Thus, solving for $$ T $$:
$$
T = \frac{\ln 2}{0.005}.
$$

Using $$ \ln 2 \approx 0.69315 $$:
$$
T \approx \frac{0.69315}{0.005} \approx 138.63 \text{ days}.
$$
**Problem 24** Initially, there were **1,250 bacteria**. **Problem 25** It will take approximately **0.59 days** for the radon to decay to 90% of its original amount. **Problem 26** The half-life of polonium-210 is approximately **138.63 days**.

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