An old bone contains 80% of its original carbon-14. use the half-life model to find the age of the bone

This is an exponential decay problem. These problems are of the form:

y = c * ekt

y = amount of substance left

c = original amount of substance (Here, set this to 100%, or 1.)

e = exponential constant (~2.718)

k = rate of decay constant (We need to figure this out.)

t = time, in years (When t = 0, y = 1 (all of the carbon-14). When t = 5730, y = 0.5 (one half-life has passed). When t = the answer we're trying to find, y = 0.98 (98% of the carbon-14).)

First, we must find the value of k, the rate of decay constant. We know that after 5730 years (t = 5730), one-half of the carbon-14 will remain (y = 0.5).

y = c * ekt

0.5 = 1 * ek * 5730

0.5 = e5730k

To get rid of the e, take the natural logarithm (ln) of both sides:

ln(0.5) = ln(e5730k)

ln(0.5) = 5730k

ln(0.5)/5730 = k

ln(0.5) is a negative number, so our rate of decay constant will be negative. This is a little "sanity check", because radioactive decay means the amount of substance goes down over time. If you get a positive value of k, then you made a mistake somewhere.

Now that we have the rate of decay constant, we can find the value of t (in years) that will yield 98% of the carbon-14 remaining.

y = c * ekt

0.98 = 1 * e[ln(0.5)/5730]t

0.98 = e[ln(0.5)/5730]t

To get rid of the e, take the natural logarithm (ln) of both sides:

ln(0.98) = ln(e[ln(0.5)/5730]t)

ln(0.98) = [ln(0.5)/5730]t

ln(0.98) = ln(0.5)t / 5730

Solve for t:

5730 * ln(0.98) = ln(0.5)t

5730 * ln(0.98) / ln(0.5) = t

Put that into your calculator, to get t ~ 167 years. Both ln(0.98) and ln(0.5) are negative, so the negatives will cancel out to yield a positive number (another "sanity check").

Step-by-step explanation:

i am thinking c. sorry if im wrong .

step-by-step explanation:

the answer is 3/2 and for decimal: 1.5

hope i

-lavira: )