Question 1196893
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a = initial value
b = determines if we have growth or decay depending if b > 1 or 0 < b < 1.


x = number of years
y = amount of carbon-14 in mg


Given info:
a = 2
b = 1 + r = 1 - 0.000124 = 0.999876
y = 0.18


y = a*b^x
0.18 = 2*0.999876^x
2*0.999876^x = 0.18
0.999876^x = 0.18/2
0.999876^x = 0.09
log(0.999876^x) = log(0.09)
x*log(0.999876) = log(0.09)
x = log(0.09)/log(0.999876)
x = 19,417.7122011154


Answer: Approximately 19,417 years old


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Here's an approach using the half-life formula


The half-life of carbon 14 is approximately H = 5730 years according to these sources here
<a href = "https://pubchem.ncbi.nlm.nih.gov/compound/carbon-14">https://pubchem.ncbi.nlm.nih.gov/compound/carbon-14</a>
<a href = "https://en.wikipedia.org/wiki/Carbon-14">https://en.wikipedia.org/wiki/Carbon-14</a>
Every 5730 years or so, the amount cuts in half.


y = a*0.5^(x/H)
0.18 = 2*0.5^(x/5730)
0.5^(x/5730) = 0.18/2
0.5^(x/5730) = 0.09
log( 0.5^(x/5730) ) = log(0.09)
(x/5730)*log(0.5) = log(0.09)
x/5730 = log(0.09)/log(0.5)
x = 5730*log(0.09)/log(0.5)
x = 19,905.6257091448
x = 19,906
This isn't too far from the 19,417 value calculated earlier. 


On the scale of tens of thousands of years, a few hundred years isn't that much of a difference 
19906-19417 = 488
489/19906 = 0.0246 = 2.46% error approximately
Though of course the level of precision will depend on what context you're in. If you're casually talking to a friend, then you don't need that much precision. For scientific papers, then you'll definitely need more accuracy.


Here's a calculator to help check your work
<a href = "https://www.omnicalculator.com/chemistry/carbon-dating">https://www.omnicalculator.com/chemistry/carbon-dating</a>
In this case there's 0.18/2 = 0.09 = 9% of the carbon 14 left
The calculator will produce the result of 19,906
This result is approximate due to the fact that the half-life 5730 was approximate.
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