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Mathematics 15 Online
OpenStudy (darkbluechocobo):

Help with Exponential Growth

OpenStudy (darkbluechocobo):

OpenStudy (darkbluechocobo):

so intial cost would be 30000?

OpenStudy (xapproachesinfinity):

hmm that easy you just plug in the values

OpenStudy (xapproachesinfinity):

yes it appears so

OpenStudy (darkbluechocobo):

so then V(12)=30000e^12b

OpenStudy (darkbluechocobo):

so the rate how would you find that?

OpenStudy (xapproachesinfinity):

hmm let me i haven't read carefully the problem

OpenStudy (xapproachesinfinity):

v(5)=v0e^(5b) solve for b

OpenStudy (xapproachesinfinity):

the v(5) is given to you

OpenStudy (xapproachesinfinity):

v0 is the initial cost which is 30000 v(5) is the cost after five years

OpenStudy (xapproachesinfinity):

the only unknown you have there is b

OpenStudy (xapproachesinfinity):

once you found b you can finally look for v(12)

OpenStudy (darkbluechocobo):

ohhhh thank you

OpenStudy (xapproachesinfinity):

eh don't use 30000 in the equation you will get a wrong answer

OpenStudy (darkbluechocobo):

so here would be 12000?

OpenStudy (xapproachesinfinity):

because the model is already in thousands

OpenStudy (xapproachesinfinity):

yeah v(5)=12 in thousands

OpenStudy (xapproachesinfinity):

do you get it

OpenStudy (darkbluechocobo):

wait so it would just be 12?

OpenStudy (xapproachesinfinity):

correct, since the model is in thousands already

OpenStudy (xapproachesinfinity):

if you use 12000 and 30000 you will get a wrong b

OpenStudy (xapproachesinfinity):

you can try and see for yourself

OpenStudy (darkbluechocobo):

V(5)=12e^5b So we have to take the natural log of both sides right?

OpenStudy (xapproachesinfinity):

no v(5)=12 remember v0=30

OpenStudy (xapproachesinfinity):

you set it the wrong way

OpenStudy (darkbluechocobo):

so just 12?

OpenStudy (darkbluechocobo):

erggg

OpenStudy (darkbluechocobo):

how do we solve for b then?

OpenStudy (xapproachesinfinity):

\(\large v(5)=v_0e^{5b} \Longrightarrow 12=30e^{5b}\)

OpenStudy (xapproachesinfinity):

do you see it now

OpenStudy (darkbluechocobo):

ohhh so basically the cost difference in the 5 years

OpenStudy (darkbluechocobo):

so would we divide 30 by both sides to start?

OpenStudy (darkbluechocobo):

chu there?

OpenStudy (xapproachesinfinity):

yes

OpenStudy (xapproachesinfinity):

sorry i was away

OpenStudy (darkbluechocobo):

aha its fine :p

OpenStudy (xapproachesinfinity):

that is easy to solve now! just divide both side by 30 reduce the lHS fraction and apply ln

OpenStudy (darkbluechocobo):

ok so .4=e^5b

OpenStudy (darkbluechocobo):

now about this how would you go about getting the exponent alone

OpenStudy (xapproachesinfinity):

eh reduce the fraction i said not a decimal

OpenStudy (xapproachesinfinity):

never go to decimal is they didn't ask you

OpenStudy (darkbluechocobo):

ohhh sorry it says nearest hundred :p

OpenStudy (xapproachesinfinity):

that would be b not that fraction

OpenStudy (darkbluechocobo):

ohhh

OpenStudy (xapproachesinfinity):

you will get a slightly wrong answer if you round before you reach the answer

OpenStudy (darkbluechocobo):

but should i keep .4 though

OpenStudy (xapproachesinfinity):

hmm eh if you want to use then go for it! my way is to just reduce the fraction and then apply ln

OpenStudy (xapproachesinfinity):

2/5 is the reduced fraction

OpenStudy (darkbluechocobo):

alright

OpenStudy (xapproachesinfinity):

\(\Large 2/5=e^{5b} \Longrightarrow \ln(2/5)=5b\)

OpenStudy (xapproachesinfinity):

well it appears that won't change anything in this case just go ahead and use 0.4

OpenStudy (darkbluechocobo):

yeh i just saw that lool

OpenStudy (darkbluechocobo):

-.91629073

OpenStudy (xapproachesinfinity):

hmm divide by 5

OpenStudy (xapproachesinfinity):

and round the nearest hundred

OpenStudy (darkbluechocobo):

-.18

OpenStudy (xapproachesinfinity):

hmm sounds good

OpenStudy (darkbluechocobo):

so now we fill this in for b in the 12 year equation

OpenStudy (xapproachesinfinity):

-18/10=-9/5=b

OpenStudy (xapproachesinfinity):

yes!

OpenStudy (xapproachesinfinity):

use b to find v(t=12)

OpenStudy (darkbluechocobo):

oks V(12)=30000e^12(-.18)

OpenStudy (xapproachesinfinity):

I said use 30 only for 30000

OpenStudy (darkbluechocobo):

Yes i forgot T_T

OpenStudy (darkbluechocobo):

oopsie

OpenStudy (xapproachesinfinity):

that's okay

OpenStudy (darkbluechocobo):

V(12)=30e^12(-.18) fixed

OpenStudy (xapproachesinfinity):

okay that's good! now compute that

OpenStudy (darkbluechocobo):

oks so 30(.11532512

OpenStudy (xapproachesinfinity):

just do the entire thing! it is just calculator work now

OpenStudy (darkbluechocobo):

lelel sorry i like writing out steps

OpenStudy (xapproachesinfinity):

once you finished multiply by 1000 to get the answer you are looking for

OpenStudy (darkbluechocobo):

3.46

OpenStudy (darkbluechocobo):

3460?

OpenStudy (xapproachesinfinity):

|dw:1420574019930:dw| apparently this how your function is acting

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