$ 2000 invested intrest compounded continuously. find time to grow account to 4500 @ 4.25%
College? What a subject. I will be happy to critique your thinking.
To find the time it takes for $2000 to grow to $4500 at an interest rate of 4.25%, compounded continuously, we can use the continuous compound interest formula:
A = P * e^(rt)
Where:
A = Final amount ($4500)
P = Initial principal ($2000)
e = Euler's number (approximately 2.71828)
r = Interest rate (4.25% or 0.0425 as a decimal)
t = Time (unknown variable)
Now substitute the known values into the formula:
4500 = 2000 * e^(0.0425t)
Divide both sides by 2000:
2.25 = e^(0.0425t)
Take the natural logarithm of both sides to isolate the exponent:
ln(2.25) = ln(e^(0.0425t))
The natural logarithm and the exponential function are inverses, so they cancel out:
ln(2.25) = 0.0425t
Now divide both sides by 0.0425:
t = ln(2.25) / 0.0425
Using a calculator to find the natural logarithm of 2.25 and the division, we get:
t ≈ 12.76
Therefore, it takes approximately 12.76 years for an initial investment of $2000 to grow to $4500 at 4.25% interest, compounded continuously.
To find the time required to grow an investment from $2000 to $4500 with continuous compounding at an interest rate of 4.25%, we can use the formula:
A = P * e^(rt)
Where:
A is the final amount
P is the initial principal
e is the base of the natural logarithm (approximately 2.71828)
r is the interest rate (in decimal form)
t is the time (in years)
In this case, A = $4500, P = $2000, r = 0.0425, and we need to solve for t.
Rearranging the formula, we get:
e^(rt) = A / P
Taking the natural logarithm (ln) of both sides, we have:
ln(e^(rt)) = ln(A/P)
rt * ln(e) = ln(A/P)
rt = ln(A/P)
Substituting the known values, we have:
t = ln(A/P) / r
Calculating the values, we get:
t = ln(4500/2000) / 0.0425 ≈ ln(2.25) / 0.0425 ≈ 3.416 / 0.0425 ≈ 80.5 years
Therefore, it would take approximately 80.5 years for the investment to grow from $2000 to $4500 with continuous compounding at an interest rate of 4.25%.