SOLUTION: the half-life of nitrogen-16 is 7 seconds. how long does it take for 100mg of nitrogen-16 be reduced to 6.25mg?

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Question 707616: the half-life of nitrogen-16 is 7 seconds. how long does it take for 100mg of nitrogen-16 be reduced to 6.25mg?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This is an exponential growth/decay problem. A general equation for these is:
A+=+A%5B0%5D%2Ar%5Et
where
t = number of units of time
A = the amount after t units of time
A%5B0%5D = the initial amount (IOW: The amount at t = 0)
r = the factor of change for 1 unit of time. If r > 1 then the equation is for growth and if 0 < r < 1 then the equation is for decay.

In your problem:
t = the number of units of 7 seconds that have passed.
A = 6.25
r = 1/2 (since the amount decreases by 1/2 every 7 seconds)
A%5B0%5D = 100
So your equation is:
6.25+=+100%2A%281%2F2%29%5Et

Now we solve for t. We start by isolating the base, 1/2, and its exponent by dividing each side by 100:
6.25%2F100+=+%281%2F2%29%5Et
which simplifies to:
0.0625+=+%281%2F2%29%5Et
If the left side is a power of 1/2 then we can solve this by hand. If not, then we will need to use exponents. It will easier to see if the left side is a power of 1/2 if we rewrite it as a fraction:
625%2F10000+=+%281%2F2%29%5Et
Clearly 25 will go into both 625 and 10000:
25%2F400+=+%281%2F2%29%5Et
Clearly 25 will go again into both 25 and 400:
1%2F16+=+%281%2F2%29%5Et
If we know (or try out) some powers of 1/2 we should quickly find that 1%2F16+=+%281%2F2%29%5E4. So t = 4. Now remember that t is how many sets of 7 seconds that have passed. So the amount reduces from 100 mg to 6.25 mg not in 4 seconds but in 4*7 or 28 seconds.