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)

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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)

Mathematics
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225 is a constant, you can pull that aside and focus on 1.15^t
ln1.15(x) = t try to substitute that
not ln...but log1.15(x) = t

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why did you take though
?
log ...im sorry
just thought itd be easier to play with.... I got no objection to another way :)
oh ok...is there a general pattern that i should follow...or even an order
i think I recall the base other than e stuff now...
a pattern..... nothing concrete that I am aware of, just find a function that you can work with easily
1.15^t is the derivative of some F(x) what do we know about exponent derivatives?
OMG im so confused
Dx(5^x) = Dx 5^x right? or am i forgetting something there...which I am prone to do
you want to integrate this function right?
yea
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?
oh ok
we dont want to change any value to it, just the way it "looks"
alright... so its log 1.15(x)= t {where did u get x from}
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"
lol..ok
lets try to "derive" a function down to something that looks close to our problem, that might help us "remember" thru the cobwebs :)
y = 5^x how would we "derive" this? do you recall?
lnx (5^x)
you sure? cause it looks right to me...
haha.. i think ...wait let me check
remember the steps to get it?
what steps?
it actually is ln(5) 5^x
thats better :)
so damn close
ok..now were ready lol
thanks for not bailing on me half way
so the derivative of 5^t would be?
i got nowhere to go ....
ln(5)5^t
and that is our "key" to this problem
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?
14 times what = 14?
*1
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?
omg...idk
lol ..... what do you know that equals 1?
does 4/2 = 1? does 4/3 =1? does 4/4 =1??
ohhh ....4/4?
instead of a "4" we need what..... ln(1.15) right?
yea
good, lets set up out integrand then...
this is our starting point \[\int\limits_{} 225(1.15^t)\]
dont we have to use F(b)-F(a)
lets pull out the constants to get this:\[225 \int\limits_{} 1.15^t\]
we need to find F(x) before we can use it lol
dang it
now lets multiply our integrand by our convenient form of "1"
\[225 \int\limits_{} [\ln(1.15)/\ln(1.15)] 1.15^t\]
tell me... is ln(1.15) a constant??
yes
then lets pull the bottom one out and leave the top in to use..sound good?
\[225/\ln(1.15) \int\limits_{} \ln(1.15) 1.15^t\]
okay that kinda makes sense
now what do we know our integrand will become?
oh shoot ......1.15^t
thats correct :) \[F(t) = [225/\ln(1.15)] 1.15^t \]
now use your "t" values and figure out the answer :)
ok...i had one more question....so my values are gonna 1 and 28?
225 ------- * 1.15^t = F(t) ln(1.15)
id subtract 1965 from both years to get a span of time
yeah i subtracted 1965 from 1983
1983?? or 1993?
1993...my bad
it aint 1 to 28, its.... zero to 28. 1965-1965=0
thats where i was confused .
1.15^0 = 1 so that aint a biggy
yeah....it was the subtraction that confused ya :) lol
OMG...i got it right...THANK YOU
80599.643 - 1609.88 =?
yay!!! that means im smart ;)
we..were smart
yea...u my friend are the man...
just dont ask me the hard questions :)
lol...its all good

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