## 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 1965-1993 (INTEGRAL PROBLEM)

1. amistre64

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

2. amistre64

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

3. amistre64

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

4. anonymous

why did you take though

5. amistre64

?

6. anonymous

log ...im sorry

7. amistre64

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

8. anonymous

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

9. amistre64

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

10. amistre64

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

11. amistre64

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

12. anonymous

OMG im so confused

13. amistre64

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

14. amistre64

you want to integrate this function right?

15. anonymous

yea

16. 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?

17. anonymous

oh ok

18. amistre64

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

19. anonymous

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

20. 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"

21. anonymous

lol..ok

22. amistre64

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

23. amistre64

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

24. anonymous

lnx (5^x)

25. amistre64

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

26. anonymous

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

27. amistre64

remember the steps to get it?

28. anonymous

what steps?

29. anonymous

it actually is ln(5) 5^x

30. amistre64

thats better :)

31. anonymous

so damn close

32. amistre64

33. anonymous

thanks for not bailing on me half way

34. amistre64

so the derivative of 5^t would be?

35. amistre64

i got nowhere to go ....

36. anonymous

ln(5)5^t

37. amistre64

and that is our "key" to this problem

38. 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?

39. amistre64

14 times what = 14?

40. anonymous

*1

41. 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?

42. anonymous

omg...idk

43. amistre64

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

44. amistre64

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

45. anonymous

ohhh ....4/4?

46. amistre64

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

47. anonymous

yea

48. amistre64

good, lets set up out integrand then...

49. amistre64

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

50. anonymous

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

51. amistre64

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

52. amistre64

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

53. anonymous

dang it

54. amistre64

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

55. amistre64

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

56. amistre64

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

57. anonymous

yes

58. amistre64

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

59. amistre64

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

60. anonymous

okay that kinda makes sense

61. amistre64

now what do we know our integrand will become?

62. anonymous

oh shoot ......1.15^t

63. amistre64

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

64. amistre64

65. anonymous

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

66. amistre64

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

67. amistre64

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

68. anonymous

yeah i subtracted 1965 from 1983

69. amistre64

1983?? or 1993?

70. anonymous

71. amistre64

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

72. anonymous

thats where i was confused .

73. amistre64

1.15^0 = 1 so that aint a biggy

74. amistre64

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

75. anonymous

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

76. amistre64

80599.643 - 1609.88 =?

77. amistre64

yay!!! that means im smart ;)

78. amistre64

we..were smart

79. anonymous

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

80. amistre64

just dont ask me the hard questions :)

81. anonymous

lol...its all good