# Chemistry students help meeee

Hey, i need a help with this university question it needs to be done ASAP.
Question 4
The half-life for the radioactive decay of 14C is 5730 years. An archaeological artifact containing wood
had only 63% of the 14C found in a living tree. Estimate the age of the sample
let x be the age of the sample:
0.5^(x/5730) = 0.63
use logarithms to solve for x
let x be the age of the sample:
0.5^(x/5730) = 0.63
use logarithms to solve for x

sorry, could you give me a step by step. i dont get it xx
Original post by welove_edu1
sorry, could you give me a step by step. i dont get it xx

of course! i was on the bus when i saw this which is why i was so brief.

because the curve of the quantity of 14C halves every given period of years, it can be expressed as y = 0.5something times x, where y is the proportion left, and x is the years

because the half-life is 5,730 years, to make that curve of y = 0.5something times x half specifically every 5,730 years, that something in the exponent has to be 1 over 5,730. this is to make it so when x is 0, y = 0.50 = 1, when x is 5,730, y = 0.51 = 0.5, when x is 11,460, y = 0.52 = 0.25 etc...

now, since you know the proportion but not the year, you can substitute y for 0.63, and solve for x:

0.5x/5,730 = 0.63
log0.5(0.5x/5,730) = log0.50.63
x/5,730 = log0.50.63
x = 5,730log0.50.63 = 3,819 years (to the nearest year)

hope this helps (:
Ermmm i still don’t get it
Original post by welove_edu1
Ermmm i still don’t get it

which part are you struggling with?
which part are you struggling with?

Oh my I thought I replied again later saying Thank you sooooo much I got it. God bless youuuu!!!
Original post by welove_edu1
Oh my I thought I replied again later saying Thank you sooooo much I got it. God bless youuuu!!!

you're welcome!! (: what's it like studying chemistry at uni?