To determine how long it will take for Derek to accumulate $2,076,331.00 by depositing $11,118.00 annually into his retirement account at an annual interest rate of 10%, we can use the future value of an annuity formula. The formula for the future value of an annuity is:
\[ FV = P \times \frac{(1 + r)^n - 1}{r} \]
Where:
- \( FV \) = future value of the annuity (the amount he wants to accumulate, $2,076,331.00)
- \( P \) = annual deposit (the amount he deposits each year, $11,118.00)
- \( r \) = annual interest rate (10% or 0.10)
- \( n \) = number of years (the variable we want to solve for)
Plugging in the values we have:
\[ 2,076,331.00 = 11,118.00 \times \frac{(1 + 0.10)^n - 1}{0.10} \]
Now, let's isolate \((1 + 0.10)^n - 1\):
\[ \frac{2,076,331.00}{11,118.00} = \frac{(1.10)^n - 1}{0.10} \]
Calculating the left side:
\[ \frac{2,076,331.00}{11,118.00} \approx 186.1 \]
Now, we rewrite the equation:
\[ 186.1 = \frac{(1.10)^n - 1}{0.10} \]
Multiplying both sides by 0.10 gives:
\[ 18.61 = (1.10)^n - 1 \]
Adding 1 to both sides:
\[ 19.61 = (1.10)^n \]
Now, to solve for \( n \), we take the logarithm of both sides:
\[ n \cdot \log(1.10) = \log(19.61) \]
Thus,
\[ n = \frac{\log(19.61)}{\log(1.10)} \]
Calculating each logarithm:
- \(\log(19.61) \approx 1.292\)
- \(\log(1.10) \approx 0.0414\)
Now:
\[ n = \frac{1.292}{0.0414} \approx 31.23 \]
Since \( n \) must be a whole number (as he makes deposits at the end of each year), we round up to the next whole number, giving \( n = 32 \).
Thus, it will take Derek approximately 32 years to accumulate $2,076,331.00 by making annual deposits of $11,118.00 at an interest rate of 10%.