The value, V , of a Tiffany lamp, worth $225 in 1965, increases at 15% per year. Its value in dollars t years after 1965 is given by
V=225(1.15)^t
Find the average value of the lamp over the period 1965-1993
(INTEGRAL PROBLEM)

- anonymous

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- amistre64

225 is a constant, you can pull that aside and focus on 1.15^t

- amistre64

ln1.15(x) = t try to substitute that

- amistre64

not ln...but log1.15(x) = t

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## More answers

- anonymous

why did you take though

- amistre64

?

- anonymous

log ...im sorry

- amistre64

just thought itd be easier to play with.... I got no objection to another way :)

- anonymous

oh ok...is there a general pattern that i should follow...or even an order

- amistre64

i think I recall the base other than e stuff now...

- amistre64

a pattern..... nothing concrete that I am aware of, just find a function that you can work with easily

- amistre64

1.15^t is the derivative of some F(x) what do we know about exponent derivatives?

- anonymous

OMG im so confused

- amistre64

Dx(5^x) = Dx 5^x right? or am i forgetting something there...which I am prone to do

- amistre64

you want to integrate this function right?

- anonymous

yea

- amistre64

then we need to "play" around with it to get it into some form that we can work with or recognize that is easy for us right?

- anonymous

oh ok

- amistre64

we dont want to change any value to it, just the way it "looks"

- anonymous

alright... so its log 1.15(x)= t {where did u get x from}

- amistre64

from the garbage can inthe back of my memory :) i think we should try it another way... I had forgotten about the derivatives or exponential functions other then base "e"

- anonymous

lol..ok

- amistre64

lets try to "derive" a function down to something that looks close to our problem, that might help us "remember" thru the cobwebs :)

- amistre64

y = 5^x how would we "derive" this? do you recall?

- anonymous

lnx (5^x)

- amistre64

you sure? cause it looks right to me...

- anonymous

haha.. i think ...wait let me check

- amistre64

remember the steps to get it?

- anonymous

what steps?

- anonymous

it actually is ln(5) 5^x

- amistre64

thats better :)

- anonymous

so damn close

- amistre64

ok..now were ready lol

- anonymous

thanks for not bailing on me half way

- amistre64

so the derivative of 5^t would be?

- amistre64

i got nowhere to go ....

- anonymous

ln(5)5^t

- amistre64

and that is our "key" to this problem

- amistre64

if we can get it to "look" like ln(1.15) 1.15^t we can easily integrate it back up right? so how do we change the way something looks without cahnge the value of it?

- amistre64

14 times what = 14?

- anonymous

*1

- amistre64

i knew that number would be useful :) we need a convenient form of the number "1". that has ln(1.15) in it...what would that be?

- anonymous

omg...idk

- amistre64

lol ..... what do you know that equals 1?

- amistre64

does 4/2 = 1?
does 4/3 =1?
does 4/4 =1??

- anonymous

ohhh ....4/4?

- amistre64

instead of a "4" we need what..... ln(1.15) right?

- anonymous

yea

- amistre64

good, lets set up out integrand then...

- amistre64

this is our starting point
\[\int\limits_{} 225(1.15^t)\]

- anonymous

dont we have to use F(b)-F(a)

- amistre64

lets pull out the constants to get this:\[225 \int\limits_{} 1.15^t\]

- amistre64

we need to find F(x) before we can use it lol

- anonymous

dang it

- amistre64

now lets multiply our integrand by our convenient form of "1"

- amistre64

\[225 \int\limits_{} [\ln(1.15)/\ln(1.15)] 1.15^t\]

- amistre64

tell me... is ln(1.15) a constant??

- anonymous

yes

- amistre64

then lets pull the bottom one out and leave the top in to use..sound good?

- amistre64

\[225/\ln(1.15) \int\limits_{} \ln(1.15) 1.15^t\]

- anonymous

okay that kinda makes sense

- amistre64

now what do we know our integrand will become?

- anonymous

oh shoot ......1.15^t

- amistre64

thats correct :)
\[F(t) = [225/\ln(1.15)] 1.15^t \]

- amistre64

now use your "t" values and figure out the answer :)

- anonymous

ok...i had one more question....so my values are gonna 1 and 28?

- amistre64

225
------- * 1.15^t = F(t)
ln(1.15)

- amistre64

id subtract 1965 from both years to get a span of time

- anonymous

yeah i subtracted 1965 from 1983

- amistre64

1983?? or 1993?

- anonymous

1993...my bad

- amistre64

it aint 1 to 28, its.... zero to 28. 1965-1965=0

- anonymous

thats where i was confused .

- amistre64

1.15^0 = 1 so that aint a biggy

- amistre64

yeah....it was the subtraction that confused ya :) lol

- anonymous

OMG...i got it right...THANK YOU

- amistre64

80599.643 - 1609.88 =?

- amistre64

yay!!! that means im smart ;)

- amistre64

we..were smart

- anonymous

yea...u my friend are the man...

- amistre64

just dont ask me the hard questions :)

- anonymous

lol...its all good

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