Math problem, constant e.
You are currently reading a thread in /wsr/ - Worksafe Requests
We respect your right to privacy. You can choose not to allow some types of cookies. Your cookie preferences will apply across our website.
The pic describes the problem.
The answer sheet says the answer is (year) 2036, but I don't know how to figure this out.
>>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.