Suppose the amount of ozone in the atmosphere is decaying exponentially, losing 0.25% each year. How many years does it take for the ozone level to drop 24%? A) 78 years B) 86 years C) 94 years D) 102 years E) 110 years

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Suppose the amount of ozone in the atmosphere is decaying exponentially, losing 0.25% each year. How many years does it take for the ozone level to drop 24%? A) 78 years B) 86 years C) 94 years D) 102 years E) 110 years

Mathematics
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Is this a first order reaction?
these problems are easier than it looks like
let's say N is the current amount and N_0 be the original amount

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Other answers:

in one year, N-0 becomes 99.75% of N_0 so \[N_1 = .9975*N_0\]
next year, N_1 becomes 99.75% of N_1 so \[N_2 = .9975*N_1\]
if you substitute N_1 with N_0, \[N_2 = (.9975)*(.9975N_0)\]
as you can see, the number of years will just tell you how many times the initial amount decreased by .25%, so the general eqn after t years is \[N_t = N_0(.9975)^t\]
I f we used this eqn, we can say that N_t = .24N_0, because it is 24% of the original state. or in other words, \[.24 = (.9975)^t\]
all you have to do is to solve for t :)
oops, sorry not .24 it is actually .76 because the wording says "drops 24%" not "is 24%" so you are solving for \[.76 = (.9925)^t\]
Is the answer 109.64 years?
since\[t = {\ln(.76) \over \ln(.9975)} \] which is approximately 109.6 I'd use 110
yep
thanks!
np :) most exponential decays and growths work the same way, so let me know if you need more help ok ?

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