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### Problem 8: Radioactive Substance Half-Life
The problem states that a radioactive substance loses \( \frac{2}{3} \) of its mass in 1 year. We are asked to find the half-life of this substance.

#### Solution:
The decay process can be modeled using the exponential decay formula:

\[
m(t) = m_0 e^{-kt}
\]

where:
- \( m(t) \) is the mass remaining after time \( t \),
- \( m_0 \) is the initial mass,
- \( k \) is the decay constant,
- \( t \) is time.

Since it loses \( \frac{2}{3} \) of its mass in 1 year, \( \frac{1}{3} \) of the mass remains after 1 year. Thus, we can write:

\[
\frac{1}{3} = e^{-k(1)}
\]

Taking the natural logarithm of both sides:

\[
\ln\left(\frac{1}{3}\right) = -k
\]

Now solve for \( k \):

\[
k = -\ln\left(\frac{1}{3}\right) = \ln(3)
\]

The half-life is the time \( t_{1/2} \) at which half of the substance remains. Thus, we set:

\[
\frac{1}{2} = e^{-kt_{1/2}}
\]

Taking the natural logarithm:

\[
\ln\left(\frac{1}{2}\right) = -k t_{1/2}
\]

Substitute \( k = \ln(3) \):

\[
\ln\left(\frac{1}{2}\right) = -\ln(3) \cdot t_{1/2}
\]

Solve for \( t_{1/2} \):

\[
t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-\ln(3)} = \frac{\ln(2)}{\ln(3)}
\]

Using approximations \( \ln(2) \approx 0.693 \) and \( \ln(3) \approx 1.099 \):

\[
t_{1/2} \approx \frac{0.693}{1.099} \approx 0.63 \text{ years}
\]

So, the half-life is approximately **0.63 years**.

---

### Problem 9: Bacterial Growth
The problem states that a bacterial culture starts with 100 bacteria, and after 2 hours the population is 250 bacteria. We are to find the number of bacteria after 4 hours, assuming exponential growth.

#### Solution:
The exponential growth model is:

\[
P(t) = P_0 e^{kt}
\]

where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 = 100 \) is the initial population,
- \( k \) is the growth constant,
- \( t \) is time.

We know that after 2 hours (\( t = 2 \)), the population is 250. Thus:

\[
250 = 100 e^{2k}
\]

Solve for \( k \):

\[
\frac{250}{100} = e^{2k}
\]
\[
2.5 = e^{2k}
\]

Taking the natural logarithm of both sides:

\[
\ln(2.5) = 2k
\]
\[
k = \frac{\ln(2.5)}{2}
\]

Now calculate \( \ln(2.5) \approx 0.916 \):

\[
k \approx \frac{0.916}{2} \approx 0.458
\]

Next, find the population after 4 hours (\( t = 4 \)):

\[
P(4) = 100 e^{4k} = 100 e^{4(0.458)}
\]

\[
P(4) = 100 e^{1.832} \approx 100 \times 6.25 = 625
\]

So, the population after 4 hours is approximately **625 bacteria**.

---

Would you like further clarifications or details?

### 5 Related Questions:
1. How do we derive the formula for exponential decay?
2. What is the difference between half-life and decay constant in radioactive decay?
3. Can exponential growth apply to real-life populations with limited resources?
4. What are other real-world examples of exponential decay?
5. How can we extend the bacterial growth problem to account for factors like resource limits?

### Tip:
When dealing with exponential growth or decay, understanding how to manipulate logarithmic functions is key to solving the equations effectively!

8. A radioactive substance lost 2/3 of its mass in 1 year. Find its half-life (in years).
9. A bacterial culture starts with 100 bacteria, and after two hours the population is 250 bacteria. Assuming that the rate of growth of the population is proportional to its size, find the number of bacteria after 4 hours.

Solve problems involving the half-life of a radioactive substance losing 2/3 of its mass in 1 year, and find the number of bacteria after 4 hours in a culture that grows exponentially. Detailed steps provided for solving using exponential decay and growth models.

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