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\[y=ne^{kt}\]

n is beginning
y is end

Correct. Dont know where to start on this problem.

so in beginning there was 400

so plug in 400 for n

plug in what for y?

5000?

yes

\[5000=400 e^{kt}\]

t represent time , what time do you see there

t would go in as 1.5

1 1/4= 1.25

I knew that.

let's go with that

k. so the equation is
\[5000=400e ^{k(1.25)}\]

yes

your job is to find k

Hmm...take ln of both sides and bring exponent down then solve for k?

exactly

\[\ln(5000)=k(1.25)\ln400e\]

not what I would do

How would you set it up?

I would divide both sides by 400 first

Ohh okay.

\[5000/400=e ^{k(1.25)}\]

now take ln of both sides

\[\ln(25/2)=k(1.25)lne\]

Wouldnt ln and e cancel eachother out though?

yes

So basically, to get k, you would divide ln(25/2) by 1.25?

yes

k=2.02?

yes

lol thanks. so to rewrite the equation, i would just plug everything in?

\[5000=400e ^{2.02(1.25)}\]

close but to be more general you use t instead of specific time

Ohh okay. So intead of 1.25, just t?

\[y=400e ^{2.02(t)}\]

that way you can find y for any t value you plug in

So on b, you would plug n as 5000 and y as 1500?

*15000

exactly , this time you are finding the 't'

I got 0.5438

Because you divide by 5000 on both sides.
Getting:
\[\ln(3)=(2.02)\ln(e)\]

And then divide by ln(2.02)
Did I do it right?

yes

Thank God. lol. But would that be like 5 mins?

what?

.5438

yes ,

it will take .5438 hours

okie dokes.

Thanks youuuu!!

nice to meet you

Nice to meet you too. It was good working with you.