SOLUTION: Please help, I am stucked at this question: Find the interest rate needed for an investment of $3,000 to grow to an amount of $4,000 in 3 years if interest is compounded contin

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Question 1025819: Please help, I am stucked at this question:
Find the interest rate needed for an investment of $3,000 to grow to an amount of $4,000 in 3 years if interest is compounded continuously. Please round the answer to the nearest hundredth of percent.
I am using the compounded interest formula:
A = P ( 1+〖r/m)〗^mt
where A = $4000 and P = $3000 and t = 3 (years) and m = 1 (annually) and r = interest (percentage)
I am stuck at this step: $4000 = $3000 〖(1+r)〗^3
Suppose if this is correct up to this point, can anyone be kind enough to show me the workings in solving this? Much thanks.
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
I am stuck at this step: $4000 = $3000 * (1+r)^3.

divide both sides of this equation by 3000 to get:

4000/3000 = (1+r)^3

take the third root of both sides of this equation to get:

(4/3)^(1/3) = 1+r

subtract 1 from both sides of this equation to get:

(4/3)^(1/3) - 1 = r

solve for r to get:

r = .1006424163.

multiply by 100 to get r = 10.06424163%.

round to the nearest tenth to get r = 10.1%

HOWEVER, .....

this would not be correct.

you needed to use the continuous compounding formula.

that formula is f = p * e^(nr)

f = future value
p = present value
e = scientific constant of 2.71828...
n = number of time periods.
r = interest rate per time period.

that formula would get you:

4000 = 3000 * e^(3r)

divide both sides of this equation by 3000 to get:

4000/3000 = e^(3r)

simplify to get:

4/3 = e^(3r)

take the natural log of both sides of this equation to get:

ln(4/3) = ln(e^(3r))

since ln(e^(3r)) is equal to 3r*ln(e), and since ln(e) = 1, then the equation becomes:

ln(4/3) = 3r

divide both sides of the equation by 3 to get:

ln(4/3)/3 = r

solve for r to get:

r = .0958940242

multiply by 100 to get r = 9.58940242%.

round to the nearest tenth to get r = 9.6%.

you needed the continuous compounding formula because the problem stated if interest is compounded continuously.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Please help, I am stucked at this question:
Find the interest rate needed for an investment of $3,000 to grow to an amount of $4,000 in 3 years if interest is compounded continuously. Please round the answer to the nearest hundredth of percent.
I am using the compounded interest formula:
A = P ( 1+〖r/m)〗^mt
where A = $4000 and P = $3000 and t = 3 (years) and m = 1 (annually) and r = interest (percentage)
I am stuck at this step: $4000 = $3000 〖(1+r)〗^3
Suppose if this is correct up to this point, can anyone be kind enough to show me the workings in solving this? Much thanks.
You used the formula for interest when doing REGULAR (monthly, quarterly, annually, etc.) compounding but this requires

the formula for CONTINUOUS ompounding, which is: .

You solve this for r, the interest rate 

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