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

Question

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

anonymous · Answer

Is this a first order reaction?

yuki · Answer

these problems are easier than it looks like

yuki · Answer

let's say N is the current amount and N_0 be the original amount

yuki · Answer

in one year, N-0 becomes 99.75% of N_0 so

$$N_1 = .9975*N_0$$

yuki · Answer

next year, N_1 becomes 99.75% of N_1 so

$$N_2 = .9975*N_1$$

yuki · Answer

if you substitute N_1 with N_0, $$N_2 = (.9975)*(.9975N_0)$$

yuki · Answer

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
$$N_t = N_0(.9975)^t$$

yuki · Answer

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, 

$$.24 = (.9975)^t$$

yuki · Answer

all you have to do is to solve for t :)

yuki · Answer

oops, sorry not .24

it is actually .76 because the wording says "drops 24%"
not "is 24%"

so you are solving for 

$$.76 = (.9925)^t$$

anonymous · Answer

Is the answer 109.64 years?

yuki · Answer

since$$t = {\ln(.76) \over \ln(.9975)} $$

which is approximately 109.6

I'd use 110

yuki · Answer

yep

anonymous · Answer

thanks!

yuki · Answer

np :) 

most exponential decays and growths work the same way,
so let me know if you need more help ok ?