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cgarr017
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I apologize if this is posted in the wrong forum. this may not be calculus but i would appreciate any help solving this equation
radioactive decay equation is N=N0e^-(0.693t/T(1/2))
N is the current specific activity value, N0 is the initial specific activity, t is the time or age of the element(Carbon in this case), and T(1/2) is the half-life value of the element. in this case the half life of carbon is 5,730 years.
I'm supposed to calculate "t" or age of the waters in the North Atlantic, North Pacific, and South Pacific.
it appears that the initial and current specific activity values are the same. Specific activity for North Atlantic is 0.93, North Pacific is 0.83 and South Pacific is 0.77.
So, starting off with the age of the North Atlantic i have the equation set up as 0.93=0.93e^-(0.693t/5730). Now my first thought is to take the natural log of both sides to get rid of the e, but the 0.93 coefficient is throwing me off. would that give you ln0.93=ln0.93+lne^-(0.693t/5730)?
radioactive decay equation is N=N0e^-(0.693t/T(1/2))
N is the current specific activity value, N0 is the initial specific activity, t is the time or age of the element(Carbon in this case), and T(1/2) is the half-life value of the element. in this case the half life of carbon is 5,730 years.
I'm supposed to calculate "t" or age of the waters in the North Atlantic, North Pacific, and South Pacific.
it appears that the initial and current specific activity values are the same. Specific activity for North Atlantic is 0.93, North Pacific is 0.83 and South Pacific is 0.77.
So, starting off with the age of the North Atlantic i have the equation set up as 0.93=0.93e^-(0.693t/5730). Now my first thought is to take the natural log of both sides to get rid of the e, but the 0.93 coefficient is throwing me off. would that give you ln0.93=ln0.93+lne^-(0.693t/5730)?