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Question 1175340: The formula for an investment worth with interest compounded annually is A = P(1 + i)^n, where P represents the initial investment, i is the interest rate, and A is the worth of the investment after n years.
a) Solve the formula for P. What was the initial investment of an investment worth $1000 that compounded 10% interest for 10 years?
b) Solve the formula for i. What is the interest rate of an investment whose worth went from $1000 to $1200 in 2 years?
c) Explain a method with which you could estimate how many years it would take for an investment to reach a certain worth at a certain interest rate.
d) Estimate how many years would it take an investment of $2100 at 20% interest to reach a worth of $5225?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula you show is:
A = P * (1 + i) ^ n
in this formula:
A is the future value
P is the present value
i is the interest rate per time period
n is the number of time periods
your problems will be taken in turn below:
a) Solve the formula for P. What was the initial investment of an investment worth $1000 that compounded 10% interest for 10 years?
in this problem, A is equal to 1000 and i is equal to .10 per year and n is equal to 10 years.
the interest rate is not shown as percent.
the interest rate percent divided by 100 is equal to the interest rate.
the formula becomes 1000 = P * (1 + .10) ^ 10
solve for P to get P = 1000 / (1 + .10) ^ 10 = 385.5432894.
confirm by solving for A using that value for P.
you get A = 385.5432894 * (1 + .10) ^ 10 = 1000.
the value of P is confirmed to be good.
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b) Solve the formula for i. What is the interest rate of an investment whose worth went from $1000 to $1200 in 2 years?
the same basic form of the equation is used.
A = P * (1 + i) ^ n becomes:
1200 = 1000 * (1 + i) ^ 2
divide both sides of this equation by 1000 to get:
1.2 = (1 + i) ^ 2
take the square root of both sides of the equation to get:
1.2 ^ (1/2) = 1 + i
solve for 1 + i to get:
1.095445115 = 1 + i
subtract 1 from both sides of the equation to get:
.095445115 = i
that's your answer.
confirm by replacing i in the original equation and solve for A to get:
A = 1000 * (1 + .095445115) ^ 2 = 1200.
this confirms the rate is good.
multiply the rate by 100 and you get 9.5445115%.
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c) Explain a method with which you could estimate how many years it would take for an investment to reach a certain worth at a certain interest rate.
the formula is, once again, A = P * (1 + i) ^ n
you want to solve for n.
divide both sides of the equation by P to get:
A/P = (1 + i) ^ n
take the log of both sides of the equation to get:
log(A/P) = log((1+i)^n)
by the law of logs that says log(x^a) = a * log(x), your equation becomes:
log(A/P) = n * log(1 + i)
divide both sides of the equation by log(1 + i) to get:
log(A/P) / log(1 + i) = n
that's the procedure you would take to find n.
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d) Estimate how many years would it take an investment of $2100 at 20% interest to reach a worth of $5225?
you would use the method described in (c) to find the value of n.
the basic formula is, once again, A = P * (1 + i) ^ n
the formula becomes 5225 = 2100 * (1 + .2) ^ n
divide both sides of the equation by 2100 to get:
5225/2100 = (1 + .2) ^ n
take the log of both sides of the equation to get:
log(5225/2100) = log((1 + .2)^n)
by the law of logs that says log(x^a) = a * log(x), your equation becomes:
log(5225/2100) = n * log(1.2)
divide both sides of the equation by log(1.2) to get:
log(5225/2100) / log(1.2) = n
solve for n to get:
n = 4.999504552.
confirm by replacing n with 4.999504552 in the equation and solve for A to get:
A = 2100 * (1 + .2) ^ 4.999504552 = 5225.
this confirms the value of n is good.
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