Question

(Pre-calculus) In 1987, the number of AIDS cases in the US was estimated to be 50,000. In 2005, this number was estimated to be 1,000,000. Assume the trend continued. [leave answers in terms of e and ln, etc]

a) Set up a function of the form N(t) = N0 x e^kt

b) According to your model, how many cases occurred in 2000?

c) When, according to your model, will there be 5,000,000 cases?

Answer 1

for part a) you need to solve for k
assign values for N_0 (initial number) and N(t) where t = 18 (2005 - 1987 = 18)

Answer 2

\[\rightarrow 1,000,000 = 50,000*e^{18k}\]

Answer 3

Ok, great thanks!! And then my constant should have an ln answer in it right?

Answer 4

correct... k = ln(20) / 18

Answer 5

ok great!! and then for part c, I would put: 5,000,000No =50,000ekt?

Answer 6

almost, N_0 = 50,000 so you don't need it on left hand side
then solve for t

Answer 7

Ok, got it! So get rid of the No and plug in 5,00,000 = 50,000e(ln20/18)t . e cancels out ln, so it would be 5mil=50000(20)(t/18)!

Answer 8

yep except the (t/18) remains an exponent...do Not multiply
\[e^{\ln 20 / 18 t} = (e^{\ln 20})^{t/18} = 20^{t/18}\]

Answer 9

Oh..don't multiply....could I possibly do log base 20 to both sides to get solve for t?

Answer 10

you could...i think its less confusing to stick with ln , then use log property to bring exponent out front

Answer 11

Ah okay! got it!! Thanks for all the much needed help. I'm taking a summer pre-calculus college class right now and our final is next week!! You're awesome!

Answer 12

your welcome..good luck

Answer 13

thanks!! I don't want to call you dumbcow because you're definitely way better than that!