In an ore containing uranium, the ratio of ^238U` to `206Pb` nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of `^238U`. Take the half-life of `^238U` to be `4.5xx10^9` years.

Explanation

To calculate the age of the ore containing uranium, we can use the relationship between the number of parent nuclei (Uranium-238) and the number of daughter nuclei (Lead-206) produced through radioactive decay.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The ratio of \( ^{238}U \) to \( ^{206}Pb \) is given as 3.
- This means that for every 3 nuclei of \( ^{238}U \), there is 1 nucleus of \( ^{206}Pb \).
- The half-life of \( ^{238}U \) is \( 4.5 \times 10^9 \) years.

2. **Set Up the Relationship:**
- Let \( N_0 \) be the initial number of \( ^{238}U \) nuclei.
- After some time \( t \), the number of \( ^{238}U \) nuclei will be \( N = N_0 e^{-\lambda t} \), where \( \lambda \) is the decay constant.
- The number of \( ^{206}Pb \) nuclei produced will be \( N_{Pb} = N_0 - N \).

3. **Express the Ratio:**
- The ratio of \( ^{238}U \) to \( ^{206}Pb \) can be expressed as:
\[ \frac{N}{N_{Pb}} = \frac{N_0 e^{-\lambda t}}{N_0 - N_0 e^{-\lambda t}} = \frac{e^{-\lambda t}}{N_0(1 - e^{-\lambda t})} \]
- Given that this ratio is 3, we can write:
\[ \frac{e^{-\lambda t}}{N_0(1 - e^{-\lambda t})} = \frac{1}{3} \]

4. **Rearranging the Equation:**
- Cross-multiplying gives:
\[ 3 e^{-\lambda t} = N_0(1 - e^{-\lambda t}) \]
- Rearranging leads to:
\[ N_0 = \frac{3 e^{-\lambda t}}{1 - e^{-\lambda t}} \]

5. **Using the Decay Constant:**
- The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9 \text{ years}} \]

6. **Substituting \( \lambda \):**
- Substitute \( \lambda \) into the earlier equation:
\[ 3 e^{-\lambda t} = N_0(1 - e^{-\lambda t}) \]
- This can be simplified to find \( e^{-\lambda t} \).

7. **Solving for \( t \):**
- From the ratio, we can derive:
\[ 4 e^{-\lambda t} = 3 \implies e^{-\lambda t} = \frac{3}{4} \]
- Taking the natural logarithm:
\[ -\lambda t = \ln\left(\frac{3}{4}\right) \implies t = -\frac{\ln\left(\frac{3}{4}\right)}{\lambda} \]

8. **Calculating \( t \):**
- Substitute \( \lambda \):
\[ t = -\frac{\ln\left(\frac{3}{4}\right)}{\frac{0.693}{4.5 \times 10^9}} \]
- Evaluating \( \ln\left(\frac{3}{4}\right) \):
\[ t \approx \frac{4.5 \times 10^9}{0.693} \cdot \ln\left(\frac{4}{3}\right) \]
- Finally, calculating gives:
\[ t \approx 1.87 \times 10^9 \text{ years} \]

### Final Answer:
The age of the ore is approximately \( 1.87 \times 10^9 \) years.