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a piece of charcoal is found to contain 30% of the carbon 14 that it originally had. when did the tree die from which the charcoal came? use 5600 years as the half-life of carbon 14.
A(t)=Aoe^kt k<0

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MissPhiladelphia
Using the equation of half-life, the given is0.5 = e^(5600k) 
in order to solve for k

we then get the log for both sides 
log(0.5) = log(e^(5600k))

then we simplify

log(0.5) = 5600k * log(e) 

k = log(0.5)/(5600*log(e)) 

k = -0.301029996 / (5600*(.434294482)

k = -0.000123776 

Back to the original equation, we have
0.3 = 1*e^(k*t)

where we substitute for k
0.3 = e^(-.000123776*t) 

and then get the log for each
log(0.3) = log(e^(-.000123776*t))

and then simplify

log(0.3) = -.000123776*t*log(e)

and then solve for t

t = log(0.3) / (-.000123776*log(e)) 

t = 9727 years

That is when the tree died.
pstnonsonjoku

The age of the sample to be determined is 1 × 10^4 years.

We know that we can obtain the rate of decay using the formula;

k = 0.693/t1/2 where t1/2 is the half life of carbon or 5600 years

k =  0.693/5600 = 1.2 × 10^-4 y-1

Using the relation;

A(t)=Aoe^-kt

A(t) = 0.3 Ao

Hence;

0.3 Ao = Aoe^-(1.2 × 10^-4t)

0.3 = e^-(1.2 × 10^-4t)

ln(0.3) = -(1.2 × 10^-4t)

t = ln(0.3) /-1.2 × 10^-4

t = 1 × 10^4 years

Learn more about carbon 14: https://brainly.com/question/4206267

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