The half-life for thorium-227 is 18.72 days. The amount $A$ (in grams) of thorium-239 after $t$ years for a 10-gram sample is given by
\[
A(t)=10 \cdot 0.5^{\frac{t}{18.72}}
\]
How long will it take before 8 grams of thorium-227 is left in the sample? Round your answer to the hundredths place.
days

Step 1 ：We are given the formula for the amount of thorium-227 left after a certain amount of time, and we are asked to find the time it takes for the amount to reduce to 8 grams. This is a problem of solving for \(t\) in the equation \(A(t) = 8\).

Step 2 ：We can rearrange the equation to solve for \(t\): \[8 = 10 \cdot 0.5^{\frac{t}{18.72}}\]

Step 3 ：Divide both sides by 10: \[0.8 = 0.5^{\frac{t}{18.72}}\]

Step 4 ：Take the natural logarithm of both sides: \[\ln(0.8) = \frac{t}{18.72} \ln(0.5)\]

Step 5 ：Solve for \(t\): \[t = \frac{18.72 \ln(0.8)}{\ln(0.5)}\]

Step 6 ：Calculate the value of \(t\) to get approximately 6.03.

Step 7 ：Final Answer: The time it will take before 8 grams of thorium-227 is left in the sample is approximately \(\boxed{6.03}\) days.