SOLUTION: The cost per megabyte of disk storage in 1981 was $700. In 2010 the cost per megabyte of disk storage was about $.00012. Assuming that the cost per megabyte is decreasing exponent

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: The cost per megabyte of disk storage in 1981 was $700. In 2010 the cost per megabyte of disk storage was about $.00012. Assuming that the cost per megabyte is decreasing exponent      Log On


   



Question 462471: The cost per megabyte of disk storage in 1981 was $700. In 2010 the cost per megabyte of disk storage was about $.00012. Assuming that the cost per megabyte is decreasing exponentially, find an exponential model of the form
C(t)=Aekt that gives the cost per megabyte C in year t. Use 1981 for t = 0, and round k to the nearest thousandth. According to the model, in what year was the cost per megabyte of storage $.12?

Answer by nerdybill(7384) About Me  (Show Source):
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The cost per megabyte of disk storage in 1981 was $700. In 2010 the cost per megabyte of disk storage was about $.00012. Assuming that the cost per megabyte is decreasing exponentially, find an exponential model of the form
C(t)=Aekt that gives the cost per megabyte C in year t. Use 1981 for t = 0, and round k to the nearest thousandth. According to the model, in what year was the cost per megabyte of storage $.12?
.
C(t)=Ae^(kt)
C(t) is .00012
A is 700
k is what we're looking for
t is 2010-1981=29
.
.00012 = 700e^(k*29)
.00012/700 = e^(k*29)
ln(.00012/700) = k*29
ln(.00012/700)/29 = k
-0.537 = k
.
our general formula:
C(t)=700e^(-0.537t)
Set c(t) to .12 and solve for t:
.12=700e^(-0.537t)
.12/700 = e^(-0.537t)
ln(.12/700) = -0.537t
ln(.12/700)/(-0.537) = t
16.148 = t
.
1981+16.148 = 1997.148
answer: 1997