Question

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!!?

Answer 1

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)|)))`