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anonymous
 5 years ago
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 19651993
(INTEGRAL PROBLEM)
anonymous
 5 years ago
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 19651993 (INTEGRAL PROBLEM)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0225 is a constant, you can pull that aside and focus on 1.15^t

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ln1.15(x) = t try to substitute that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0not ln...but log1.15(x) = t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why did you take though

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0just thought itd be easier to play with.... I got no objection to another way :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok...is there a general pattern that i should follow...or even an order

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i think I recall the base other than e stuff now...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a pattern..... nothing concrete that I am aware of, just find a function that you can work with easily

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01.15^t is the derivative of some F(x) what do we know about exponent derivatives?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(5^x) = Dx 5^x right? or am i forgetting something there...which I am prone to do

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you want to integrate this function right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we dont want to change any value to it, just the way it "looks"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright... so its log 1.15(x)= t {where did u get x from}

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0from 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"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets try to "derive" a function down to something that looks close to our problem, that might help us "remember" thru the cobwebs :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = 5^x how would we "derive" this? do you recall?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you sure? cause it looks right to me...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha.. i think ...wait let me check

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0remember the steps to get it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it actually is ln(5) 5^x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok..now were ready lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for not bailing on me half way

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so the derivative of 5^t would be?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i got nowhere to go ....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and that is our "key" to this problem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0i 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol ..... what do you know that equals 1?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0does 4/2 = 1? does 4/3 =1? does 4/4 =1??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0instead of a "4" we need what..... ln(1.15) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good, lets set up out integrand then...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this is our starting point \[\int\limits_{} 225(1.15^t)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont we have to use F(b)F(a)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets pull out the constants to get this:\[225 \int\limits_{} 1.15^t\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we need to find F(x) before we can use it lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now lets multiply our integrand by our convenient form of "1"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[225 \int\limits_{} [\ln(1.15)/\ln(1.15)] 1.15^t\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tell me... is ln(1.15) a constant??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then lets pull the bottom one out and leave the top in to use..sound good?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[225/\ln(1.15) \int\limits_{} \ln(1.15) 1.15^t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay that kinda makes sense

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now what do we know our integrand will become?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh shoot ......1.15^t

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats correct :) \[F(t) = [225/\ln(1.15)] 1.15^t \]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now use your "t" values and figure out the answer :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...i had one more question....so my values are gonna 1 and 28?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0225  * 1.15^t = F(t) ln(1.15)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id subtract 1965 from both years to get a span of time

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i subtracted 1965 from 1983

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it aint 1 to 28, its.... zero to 28. 19651965=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats where i was confused .

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01.15^0 = 1 so that aint a biggy

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah....it was the subtraction that confused ya :) lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OMG...i got it right...THANK YOU

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.080599.643  1609.88 =?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yay!!! that means im smart ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea...u my friend are the man...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0just dont ask me the hard questions :)
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