The finction A=283e^(0.028t) models the population of a particular city, in thousands, t years after 1998. When will the city reach 335 thousand?

Question

harsimran_hs4 · Answer

simple just need to solve 283e^(0.028t) > 335 
so 
e^(0.028t) > 335/283
solve this equation...

if problem persists please ask

anonymous · Answer

Why is it greater than rather than equals?

harsimran_hs4 · Answer

oops sorry i just read the question wrong it`s equal instead of = (i tought cross 335k mark)

harsimran_hs4 · Answer

* instead of >

anonymous · Answer

Okay! (: How would you keep solving it out with the e involved?

harsimran_hs4 · Answer

to remove e take ln on both sides

dumbcow · Answer

take log of both sides
$$\ln e^{.028t} = \ln \frac{335}{283}$$
$$.028t = \ln \frac{335}{283}$$

anonymous · Answer

Aye, dumbcow, the ln wouldn't stay on the left side?

harsimran_hs4 · Answer

so what it`s constant on right side....you can find the value

anonymous · Answer

Do you get 8 years?

anonymous · Answer

Nevermind, I messed up! It's 6 years. Thanks! (:

harsimran_hs4 · Answer

yes approximately 6yrs :)