On August 8,1981 , American Savings offered an insured tax-free account paying \( 23.24 \% \) compounded monthly. If you had invested \( \$ 50,000 \) at that time, how much would you have on August 8,2014 , assuming that you could have locked the interest rate at the time deposit? Use the formula \[ A=P\left(1+\frac{r}{n}\right)^{n t} \] for \( n=12 \). (Round your answer to the nearest cent.)

We are given the formula for compound interest   A = P · (1 + r/n)^(n · t) where   P = 50,000 dollars (initial deposit)   r = 23.24% = 0.2324 (annual interest rate)   n = 12 (compounding periods per year)   t = number of years from August 8, 1981 to August 8, 2014. Since 2014 − 1981 = 33, we have t = 33 years. Step 1. Write the formula with the given values:   A = 50,000 · (1 + 0.2324/12)^(12 · 33) Step 2. Compute the periodic interest rate:   0.2324/12 ≈ 0.01936667 Then the factor is (1 + 0.01936667) ≈ 1.01936667. Step 3. Compute the number of compounding periods:   n · t = 12 · 33 = 396 So the accumulation factor becomes (1.01936667)^396. Step 4. Calculate the total amount A: While a hand‐calculation may be tricky, we can use the natural logarithm to see the size:   ln[(1.01936667)^396] = 396 · ln(1.01936667) Using the approximation ln(1.01936667) ≈ 0.01917, we have   396 · 0.01917 ≈ 7.591 Thus   (1.01936667)^396 ≈ e^(7.591) ≈ 1981 Then   A ≈ 50,000 · 1981 = 99,050,000 Step 5. Round to the nearest cent: The result is approximately   A ≈ $99,050,000.00 Thus, had you invested $50,000 at 23.24% compounded monthly, by August 8, 2014 you would have approximately $99,050,000.00.