Question 1104553: Please help me solve this word problem:
The number of households using online banking services has increased from 754,000 in 1995 to 12,980,000 in 2000. The formula H(t)=0.76e^0.55t models the number of households H in millions when time is t years since 1995. In what year was it estimated that 50,000,000 households would use online banking services?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula is not correct.
this cause problems in understanding what is going on.
the first thing you need to do is convert eveything to millions of households because the formula assumes millions of households.
therefore:
in 1995, you have .758 million households.
in 2000, you have 12.8 million households.
the exponential formula is f = p * e^(rt).
f is the future value.
p is the present value.
r is the interest rate per year.
t is the number of years.
in your problem:
f = 12.8
p = .758
r = what you want to find, since .55 doesn't appear to be right.
t = 5 years since 2000 - 1995 = 5.
formula becomes 12.8 = .758 * e^(5r)
divide both sides of this formula by .758 to get 12.8/.758 = e^(5r).
take the natural log of both sides of this equation to get ln(12.8/.758) = ln(e^(5r)).
since ln(e^(5r)) is equal to 5r * ln(e),then your formula becomes ln(12.8/.758) = 5r * ln(e).
since ln(e) is equal to 1, then your formula becomes ln(12.8/.758) = 5r
divide both sides of this formula by 5 to get ln(12.8/.758) / 5 = r
solve for r to get r = .5653034129.
that is slightly larger than .55 that you were given.
it should, consequently, result in a more accurate answer.
to test it out, we'll use r = .5653034129 rather than .55 to see what we get.
we will also use .758 rather than .76, since .758 million is more accurate than .76 million.
using .758 million in 1995 and r = .5653034129, we get the following:
in 1995, the number of households using online banking services is .758 * e^(.5653034129 * 0) which becomes .758 * e^0 which is equal to .758 million.
in 2000, the number of households using online banking services is .758 * e^(.5653034129 * 5) which is equal to 12.8 million.
since we're right on, we can be confident that the new value of r is the correct value.
the question remains:
in what year will the number of households using online banking services be equal to 50,000,000.
convert that to millions of households, and the question becomes in what year will the number of households using online banking services be equal to 50 million.
the formula is still f = p * e^(rt).
p is equal to .758.
f is equal to 50.
r is equal to .5653034129
t is what we want to find.
the formula becomes 50 = .758 * e^(.5653034129 * t)
divide both sides of the equation by .758 to get 50/.758 = e^(.5653034129 *t)
take the natural log of both sides of this equation to get ln(50/.758) = ln(e^(.5653034129 * t)).
since ln(e^(.5653034129 * t)) is equal to .5653034129 * t * ln(e), and since ln(e) = 1, then the formula becomes ln(50/.758) = .5653034129 * t
divide both sides of this formula by .5653034129 to get ln(50/.758) / .5653034129 = t.
solve for t to get t = 7.410347795 years.
the number of households using online banking services should be equal to 50 million in 7.410347795 years from 1995.
to test this out, use the formula to get .758 * e^(.5653034129 * 7.410347795) equals 50 million.
the solution is confirmed to be good.
the solution is that the number of households using online banking services will be equal to 50 million in 7.410347795 years from 1995.
that would put it somewhere between 2002 and 2003.
this means that it will occur in 2002.
the formula that you were given is not correct and that's what might have created some difficulty for you, if you were able to catch it.
if you weren't, and you used the formula as given, then you should have done the following.
formula of f = .76 * e^(.55 * t) becomes 50 = .76 * e^(.55 * t)
divide both sides of this equation by .76 to get 50/.76 = e^(.55 * t).
take natural log of both sides of this equation to get ln(50/.76) = ln(e^(.55 * t)).
since ln(e^(.55 * t) becomes .55 * t * ln(e) which becomes .55 * t, then the formula becomes ln(50/.76) = .55 * t.
divide both sides of this equation by .55 to get ln(50/.76)/ .55 = t
solve for t to get t = 7.611745184.
this puts the solution somewhere between 2002 and 2003, which results in the solution being sometime in the year 2002.
you wind up with the same answer, because the discrepancy in the interest rate wasn't large enough to make a difference in which year the 50 million occurred.
bottom line is that the year is 2002 regardless if you use the formula as given or you used the more correct interest rate per year that was calculated above.
i took the liberty of graphing both equations separately and then graphing them together to show you what the solution would look like graphically.
first graph is using r = .55 in the equation.
second graph is using r = .5653034129 in the equation.
third graph shows both equations on the same graph.
fourth graph shows that the solution is between the 7th year and the 8th year after 1995.
since the 7th year is equal to 2002, then the number of households reaches 50 million in the year 2002.
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