Math problem, constant e.

>>81745
Ok, so you're looking for the year where N>80,000. Set N = 80,000. Then you have:

80000 = 52000*e^(.012t)
Divide both sides by 52k:

80000/52000 = e^(.012t)

Take the natural log on both sides:
ln(80000/52000) = ln(e^(.012t))

Now, as you've probably learned at some point, ln(e^x) = x. So, when you reduce ln(e^(.012t)) you get:
ln(80000/52000) = .012t

Divide both sides by .012:

ln(80000/52000)/.012 = t
t=35.898576341

I'm not entirely sure why they're rounding up to 2036 for the answer, but maybe the precise wording of the question makes that clear.

>>81746
Thank you! This helped! :)

>>81746
Bc they're looking for a round year and in 2035 the population isnt big enough yet.