The half-life of carbon 14 is 5,730 years. Approximately, how old is a bone that has 70% of its original carbon 14?

To solve this problem, we need to use the formula for exponential decay which is N = N0 * (1/2)^(t/h), where:

• N is the final amount of the substance
• N0 is the initial amount of the substance
• t is the time that has passed
• h is the half-life of the substance

In this case, we know that N/N0 = 70/100 = 0.7, and h = 5730 years. We want to find t.

So, we have 0.7 = (1/2)^(t/5730).

To solve for t, we'll need to use logarithms. We can take the natural log (ln) of both sides:

ln(0.7) = (t/5730) * ln(1/2).

Now, solve for t:

t = 5730 * ln(0.7) / ln(1/2).

Using a calculator, we find that t ≈ 2556 years.

So, the bone is approximately 2556 years old.