SOLUTION: If $7000 is invested at 8% compounded monthly, how much will this amount be at the end of 15 years? Use the formula, A = P (1 + r/n)^nt,where n is the number of times the investme

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Question 601674: If $7000 is invested at 8% compounded monthly, how much will this amount be at the end of 15 years? Use the formula, A = P (1 + r/n)^nt,where n is the number of times the investment is compounded per year. How long will it be until it is worth $12000?
I need one of the awesome tutor's to help me with this word problems, as I keep struggling trying to do the set up to solve. I get confused on how to know where to plug the values into the equation formula. Please help and explain step by step the reasoning behind the solution.
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Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Part 1)
If $7000 is invested at 8% compounded monthly, how much will this amount be at the end of 15 years?



A=P%281%2Br%2Fn%29%5E%28n%2At%29 Start with the compound interest formula


A=7000%281%2B0.08%2F12%29%5E%2812%2A15%29 Plug in P=7000, r=0.08 (the decimal equivalent of 8%), n=12 (since we're compounding 12 times a year) and t=15.


A=7000%281%2B0.00666666666666667%29%5E%2812%2A15%29 Evaluate 0.08%2F12 to get 0.00666666666666667


A=7000%281.00666666666667%29%5E%2812%2A15%29 Add 1 to 0.00666666666666667 to get 1.00666666666667


A=7000%281.00666666666667%29%5E%28180%29 Multiply 12 and 15 to get 180.


A=7000%283.30692147741001%29 Evaluate %281.00666666666667%29%5E%28180%29 to get 3.30692147741001.


A=23148.45034187 Multiply 7000 and 3.30692147741001 to get 23148.45034187.


A=23148.45 Round to the nearest hundredth (ie to the nearest penny).


So at the end of 15 years, there will be $23,148.45 in the account.

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Part 2)

How long will it be until it is worth $12000?




A=P%281%2Br%2Fn%29%5E%28n%2At%29 Start with the compound interest formula


12000=7000%281%2B0.08%2F12%29%5E%2812%2At%29 Plug in A=12000, P=7000, r=0.08 (the decimal equivalent of 8%), and n=12.


12000=7000%281%2B0.00666666666666667%29%5E%2812%2At%29 Evaluate 0.08%2F12 to get 0.00666666666666667


12000=7000%281.00666666666667%29%5E%2812%2At%29 Add 1 to 0.00666666666666667 to get 1.00666666666667


12000%2F7000=%281.00666666666667%29%5E%2812%2At%29 Divide both sides by 7000.


1.71428571428571=%281.00666666666667%29%5E%2812%2At%29 Evaluate 12000%2F7000 to get 1.71428571428571.


ln%281.71428571428571%29=ln%28%281.00666666666667%29%5E%2812%2At%29%29 Take the natural log of both sides.


ln%281.71428571428571%29=12%2At%2Aln%281.00666666666667%29 Pull down the exponent using the identity log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29.


ln%281.71428571428571%29%2Fln%281.00666666666667%29=12%2At Divide both sides by ln%281.00666666666667%29.


0.538996500732687%2Fln%281.00666666666667%29=12%2At Evaluate the natural log of 1.71428571428571 to get 0.538996500732687.


0.538996500732687%2F0.00664454271866851=12%2At Evaluate the natural log of 1.00666666666667 to get 0.00664454271866851.


81.118674911717=12%2At Divide.


81.118674911717%2F12=t Divide both sides by 12 to isolate "t".


6.75988957597641=t Divide.


t=6.75988957597641 Rearrange the equation.


t=7 Round to the nearest whole number (ie to the nearest whole year).


So it will take about 7 years for the account to be worth $12,000