To solve this problem, we will use the exponential decay formula, which describes how substances like carbon-14 decay over time:
\[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}} \]
Where:
- \( N(t) \) is the quantity of the substance remaining after time \( t \)
- \( N_0 \) is the original quantity of the substance
- \( T \) is the half-life of the substance
- \( t \) is the elapsed time
In this case, \( N_0 = 10 \) grams and \( N(t) = 8 \) grams. We are trying to find \( t \). The half-life \( T \) of carbon-14 is 5730 years. So plugging in these values gives us:
\[ 8 = 10 \times \left(\frac{1}{2}\right)^{\frac{t}{5730}} \]
Next, we'll solve for \( t \). First, divide both sides by 10 to isolate the exponential term:
\[ \frac{8}{10} = \left(\frac{1}{2}\right)^{\frac{t}{5730}} \]
\[ \frac{4}{5} = \left(\frac{1}{2}\right)^{\frac{t}{5730}} \]
Now, to solve for \( t \), we can use logarithms. Starting with the previous equation, take the natural logarithm (or logarithm base 10, or any common logarithm, since we're only interested in the ratio) of both sides:
\[ \ln\left(\frac{4}{5}\right) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right) \]
Using the property of logarithms \( \ln(a^b) = b \ln(a) \), we get:
\[ \ln\left(\frac{4}{5}\right) = \frac{t}{5730} \ln\left(\frac{1}{2}\right) \]
Next, we'll isolate \( t \) by multiplying both sides of the equation by 5730 and dividing by \( \ln\left(\frac{1}{2}\right) \):
\[ t = \frac{5730 \times \ln\left(\frac{4}{5}\right)}{\ln\left(\frac{1}{2}\right)} \]
Since the natural logarithm of \( \frac{1}{2} \) is negative, when we divide by it, the sign will invert, properly reflecting the positive passage of time.
Performing this calculation, we find that it will take approximately 1844.65 years for a 10-gram sample of carbon-14 to decay to an 8-gram sample.
