Solving this equation radioactive decay equation

[cgarr017]

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)?

 

[HallsofIvy]

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)?

That "it appears that the initial and current specific activity values are the same" can't be right. If it were true, for example, that 0.93=0.93e^-(0.693t/5730), then the first thing you would do is divide both sides by 0.93 to get e^(0.693t/5730)= 1 and then t= 0.

That's because that formula: N= N0e^(-0.693t/5730) tells you that N is steadily decreasing from N0. The only time N= N0 is when t= 0. The initial and current specific activities CAN'T be the same.

 

[cgarr017]

OK, so here's the problem. tell me what you think N and N0 should be.
The youngest bottom waters in the North Atlantic had a [delta 14 C] value of approx -70o/oo (o/oo being parts per thousand). Bottom waters in the South Pacific have a [delta 14 C] value of -170o/oo, while bottom waters in the North Pacific had a [delta 14 C] value of -230o/oo.
a)In the [delta 14 C] notation, the specific sample activity is approximately given by 1+([delta 14 C]/1000). Using this fromula, convert these [delta 14 C] values to specific carbon 14 activities. For this i got north atlantic-> 1+ (-70/1000)=0.93, south pacific->0.83, north pacific->0.77.
b)Now, assume that a water mass sinks out of the north atlantic at t=0 with the specific activity based on a [delta 14 C] value of -70o/oo. Using the radioactive decay equation N=N0e^-(0.693t/(T1/2)), calculate the "age" of the waters in the south pacific and north pacific. recall that T1/2 is half-life and the half-life of carbon is 5,730 years.
c)The average replacement or residence time for the entire ocean is ~500-1000 years. Please discuss the ages you calculated for north pacific and south pacific waters in the context of this average replacement time.