Suppose a bone from an ancient human contains 16.5% of the C-14 found in living organisms. How old is the bone? Please help; thank you so much!!?

1 Answer
Mar 31, 2016

The bone is approximately 14894.92 years old.

Explanation:

When dealing with a half life question, it is best to use the half-life formula, which is expressed as:

color(blue)(|bar(ul(color(white)(a/a)y=a(b)^(t/h)color(white)(a/a)|)))

where:
y=final amount
a=inital amount
b=growth/decay
t=time elapsed
h=half-life

Note that the half-life of carbon-14 is approximately 5730 years.

1. Start by expressing 16.5% as a decimal and plugging the value into the half-life formula as well as your other known values.

  • a=1 is 100%, expressed as a decimal
  • b=1/2 indicates half-life

y=a(b)^(t/h)

0.165=1(1/2)^(t/5730)

2. Since the bases are not the same on both sides of the equation, take the logarithm of both sides.

log(0.165)=log((1/2)^(t/5730))

3. Use the logarithmic property, log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m), to simplify the right side of the equation.

log(0.165)=(t/5730)log(1/2)

4. Isolate for t.

t/5730=(log(0.165))/(log(1/2))

t==(5730log(0.165))/(log(1/2))

5. Solve.

color(green)(|bar(ul(color(white)(a/a)t~~14894.92color(white)(i)"years old"color(white)(a/a)|)))