The half-life of carbon-14 is 5600 years. If charred logs from an old log cabin show only 71% of the carbon-14 expected in living matter, when did the cabin burn down?

Assume that the cabin burned soon after it was built from freshly cut logs?

1 Answer
Johnson Z.
Mar 31, 2016

The cabin burned down approximately #2767.01# years ago.

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

#1#. Start by expressing #71%# 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)#

#0.71=1(1/2)^(t/5600)#

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

#log(0.71)=log((1/2)^(t/5600))#

#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(0.71)=(t/5600)log(1/2)#

#4#. Isolate for #t#.

#t/5600=(log(0.71))/(log(1/2))#

#t==(5600log(0.71))/(log(1/2))#

#5#. Solve.

#color(green)(|bar(ul(color(white)(a/a)t~~2767.01color(white)(i)"years old"color(white)(a/a)|)))#

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