To find the half-life of a radioactive substance, we can use the decay formula:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} \]
where:
- \( N(t) \) is the remaining quantity of the substance (3g),
- \( N_0 \) is the initial quantity (25g),
- \( t \) is the time elapsed (17527.5 years),
- \( T_{1/2} \) is the half-life, which we want to find.
Rearranging the equation to solve for \( T_{1/2} \):
\[ \frac{N(t)}{N_0} = \left( \frac{1}{2} \right)^{t/T_{1/2}} \]
Substituting the known values:
\[ \frac{3}{25} = \left( \frac{1}{2} \right)^{17527.5/T_{1/2}} \]
Calculating \( \frac{3}{25} \):
\[ \frac{3}{25} = 0.12 \]
Now we take the logarithm of both sides:
\[ \log(0.12) = \log\left( \left( \frac{1}{2} \right)^{17527.5/T_{1/2}} \right) \]
Using the property of logarithms, we can simplify the right side:
\[ \log(0.12) = \frac{17527.5}{T_{1/2}} \log(0.5) \]
Now, calculate \( \log(0.12) \) and \( \log(0.5) \):
- \( \log(0.12) \approx -0.9208 \)
- \( \log(0.5) \approx -0.3010 \)
Substituting these values into the equation:
\[ -0.9208 = \frac{17527.5}{T_{1/2}} \cdot (-0.3010) \]
Solving for \( T_{1/2} \):
\[ -0.9208 = \frac{-17527.5 \cdot 0.3010}{T_{1/2}} \]
This simplifies to:
\[ T_{1/2} = \frac{17527.5 \cdot 0.3010}{0.9208} \]
Calculating the right side:
\[ T_{1/2} \approx \frac{5282.225}{0.9208} \approx 5734.56 \]
Thus, the half-life \( T_{1/2} \) of the sample is approximately 5734.56 years.