What is the effective interest rate corresponding to 3.75% compounded continuously? How do I find this? I'm thinking about using the formula A = P(1+r/t)^t ??
you need to find r such that
(1+r) = e^0.0375
r = 0.0382 or 3.82%
To find the effective interest rate corresponding to 3.75% compounded continuously, you are on the right track. The formula you mentioned, A = P(1+r/t)^t, is used for compound interest, where A is the final amount, P is the initial principal, r is the annual interest rate, and t is the time in years. However, in the case of continuous compounding, we use a slightly different formula.
In continuous compounding, the formula is A = Pe^(rt), where e is the mathematical constant approximately equal to 2.71828.
To find the effective interest rate, you can rearrange the formula as follows:
A = Pe^(rt)
==> A/P = e^(rt)
==> ln(A/P) = rt
==> r = (1/t) * ln(A/P)
Now, let's apply this formula to find the effective interest rate.
For your case, you have an interest rate of 3.75% (or 0.0375) compounded continuously. To find the effective interest rate, you need to know the initial principal, P, as well as the time, t.
Let's say you have $1,000 as the initial principal and you want to know the effective interest rate after 1 year.
Substituting the values into the formula, you have:
r = (1/1) * ln(A/P) = ln(A/1000)
To calculate the effective interest rate, you need to know the final amount, A. Let's solve for the final amount:
A = Pe^(rt)
==> A = 1000 * e^(0.0375 * 1)
==> A ≈ 1037.007
Now substituting the values into the formula for the effective interest rate:
r = ln(A/1000)
==> r = ln(1037.007/1000)
==> r ≈ ln(1.037007)
==> r ≈ 0.036352
So, the effective interest rate corresponding to 3.75% compounded continuously after 1 year is approximately 3.6352%.
You can apply this process to any other values by adjusting the initial principal, time, and final amount accordingly.
Note: The choice of the natural logarithm (ln) is based on the fact that the exponential function used in the continuous compounding formula is based on the natural logarithm.