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Question 1203599: 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) Rearrange the formula for P. What was the initial investment of an investment worth $1000 that compounded 10% interest for 10 years?
b) Rearrange 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?
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20850)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 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) Rearrange the formula for P. What was the initial investment of an investment worth $1000 that compounded 10% interest for 10 years?
start with A = P * (1 + i) ^ n.
divide both sides of the equation by (1 + i) ^ n to get:
A / ((1 + i) ^ n) = P
switch sides to get:
P = A / ((1 + i) ^ n)
when A = 1000 and i = 10% interest copounded for 10 years, the formula becomes:
P = 1000 / ((1 + .10) ^ 10 = 385.5432894.
that's your solution.
by replacing P with that in the original equation and appling 10% compounded yearly to it for 10 years to get:
A = 385.5432894 * (1 + .10) ^ 10 = 1000.
this confirms that P was calculated correctly.
b) Rearrange the formula for i. What is the interest rate of an investment whose worth went from $1000 to $1200 in 2 years?
A = P * (1 + i) ^ n becomes:
1200 = 1000 * (1 + i) ^ 2
divide both sides of the 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
subtract 1 from both sides of the equation to get:
1.2 ^ (1/2) - 1 = i
simplify to get:
.095445115
that's your interest rate.
confirm by replacing i in the original equation and solving for A to get:
A = 1000 * (1 + .095445115) ^ 2 = 1200.
this confirms that the value of i is correct.
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 general formula is A = P * (1 + i) ^ n
you know A and you know P and you know i, but you don't know n.
you would use logarithms to find the value of n as follows.
you would divide both sides of the equation by P to get:
A/P = (1 + i) ^ n
you would take the log of both sides of this equation to get:
log(A/P) = log((1+i)^n)
by rule of logs, this becomes:
log(A/P) = n * log(1 + i)
you would divide both sides of the equation by log(1 + i) to get:
log(A/P) / log(1 + i) = n
d) Estimate how many years would it take an investment of $2100 at 20% interest to reach a worth of $5225?
A = P * (1 + i) ^ n becomes:
5225 = 2100 * (1 + .2) ^ n.
simplify to get:
5225 = 2100 * 1.2 ^ n
divie 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 rule of logs, this 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.
replace n in the oiginal equation with that and solve for A to get:
A = 2100 * 1.2 ^ 4.999504552 = 5225, confirming the value of n is good.
note that the log rule that allows this is log(b^a) = a * log(b).
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