Quick calc help

Which gives:

N(t) \;=\;3000e^{(\frac{ln(3)}{3})t}

...............................

(=\;3000\cdot3^{\frac{t}{3}})

N(t) = 9000e^{\frac{ln3}{3}t} - 4500t

Try putting N(t) = 0, and see if you can find a value of t that fits.

i said (ii) not (iii)

which math subject is this...yikes

batman_

Which gives:

N(t) \;=\;3000e^{(\frac{ln(3)}{3})t}

...............................

(=\;3000\cdot3^{\frac{t}{3}})

Without looking at all the question, this is clearly wrong.

e^{(\frac{1}{3}\ln(3))t}=e^t.e^{\ln(3^{\frac{1}{3}})}=3^{\frac{1}{3}}e^t

edit: Having looked at the question, I agree with your answer for i) and that N(t)=

3000\times 3^{\frac{1}{3}}e^t

The equation for ii) is the first one you said but with a slight difference... You have

N(t)=3000\times 3^{\frac{1}{3}}e^t-4500(t-3)

(only valid for t>3) Do you see why?

nota bene

The equation for ii) is the first one you said but with a slight difference... You have

N(t)=3000\times 3^{\frac{1}{3}}e^t-4500(t-3)

(only valid for t>3) Do you see why?

Not really, why is there a (t-3)?
(I was beginning to to think the 3rd was right...)

edit: Having looked at the question, I agree with your answer for i) and that N(t)=

3000\times 3^{\frac{1}{3}}e^t

you wrote that not me, unless u meant

N(t)=\;3000\cdot3^{\frac{t}{3}}

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