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 $\boxed{{\displaystyle 6.03}}$ days.