Question:** A mathematician working in applied mathematics at a financial institution models compound interest with a discrete-time model. Suppose an investment grows according to the recurrence \( A_{n+1} = r A_n \) with \( A_0 = P_0 \). If \( A_5 = 3P_0 \), find the annual growth rate \( r \) and the smallest integer \( n \) such that \( A_n > 10P_0 \).

["Title: Modeling Compound Interest: Finding Growth Rate and Doubling Time in a Discrete Financial Model", "Meta Description:\nExplore how compound interest is modeled using a discrete recurrence relation ( A_{n+1} = r A_n ). Given ( A_5 = 3P_0 ), we solve for the annual growth rate ( r ) and determine the smallest integer ( n ) such that the investment surpasses ( 10P_0 ).", "---", "### Introduction to Compound Interest in Applied Mathematics", "In applied mathematics, particularly within financial modeling, compound interest is a cornerstone concept. Unlike simple interest, compounding applies growth iteratively—each period’s return is calculated on the current balance, not just the principal. A classic discrete model uses the recurrence:", "[\nA_{n+1} = r A_n\n]", "where ( A_n ) is the account balance after ( n ) periods, ( r ) is the annual growth factor (growth rate), and ( A_0 = P_0 ) is the initial investment.", "Given real-world data, financial mathematicians often determine ( r ) by analyzing past growth patterns—such as investments growing to triple in 5 years—and then project future values. This article solves for the growth rate ( r ) under the condition ( A_5 = 3P_0 ), then computes the smallest ( n ) such that ( A_n > 10P_0 ).", "---", "### Step 1: Solving for the Annual Growth Rate ( r )", "The recurrence ( A_{n+1} = r A_n ) with ( A_0 = P_0 ) generates a geometric sequence:", "[\nA_n = P_0 \cdot r^n\n]", "We are told that ( A_5 = 3P_0 ). Substituting into the formula:", "[\nA_5 = P_0 \cdot r^5 = 3P_0\n]", "Divide both sides by ( P_0 ):", "[\nr^5 = 3\n]", "To solve for ( r ), take the fifth root of both sides:", "[\nr = 3^{1/5}\n]", "This is the exact annual growth factor. To express it numerically:", "[\nr = 3^{0.2} \approx 1.24573\n]", "So, the annual growth rate is ( r \approx 1.24573 ), or 24.573% per year.", "---", "### Step 2: Finding the Smallest Integer ( n ) such that ( A_n > 10P_0 )", "We now want the smallest integer ( n ) for which:", "[\nA_n > 10P_0\n]", "Using the closed-form expression:", "[\nP_0 \cdot r^n > 10P_0\n]", "Divide both sides by ( P_0 ):", "[\nr^n > 10\n]", "Substitute ( r = 3^{1/5} ):", "[\n(3^{1/5})^n = 3^{n/5} > 10\n]", "Take logarithms (base 10 or natural log) of both sides. Using base 10:", "[\n\log_{10}(3^{n/5}) > \log_{10}(10)\n]", "[\n\frac{n}{5} \log_{10}(3) > 1\n]", "We know ( \log_{10}(3) \approx 0.4771 ), so:", "[\n\frac{n}{5} \cdot 0.4771 > 1 \quad \Rightarrow \quad n > \frac{5}{0.4771} \approx 10.476\n]", "The smallest integer ( n ) satisfying this is:", "[\nn = 11\n]", "---", "### Verification", "Let’s verify ( A_{11} ) and confirm it exceeds ( 10P_0 ):", "[\nA_{11} = P_0 \cdot (3^{1/5})^{11} = P_0 \cdot 3^{11/5} = P_0 \cdot 3^{2.2}\n]", "Compute ( 3^{2.2} = 3^2 \cdot 3^{0.2} = 9 \cdot 3^{0.2} \approx 9 \cdot 1.24573 = 11.2116 )", "Thus, ( A_{11} \approx 11.21P_0 > 10P_0 ) — correct.", "Check ( A_{10} = P_0 \cdot 3^{2} = 9P_0 < 10P_0 ) — indeed, ( n = 10 ) is insufficient.", "---", "### Conclusion", "Given a financial model where an investment grows via ( A_{n+1} = r A_n ) and triples in 5 years (( A_5 = 3P_0 )), the annual growth rate is:", "[\nr = 3^{1/5} \approx 1.24573\n]", "Using this rate, the smallest integer ( n ) for which the investment exceeds 10 times the initial principal ( P_0 ) is:", "[\nn = 11\n]", "This illustrates the power of exponential models in forecasting long-term financial growth—an essential tool for applied mathematicians in quantitative finance.", "---", "Keywords: compound interest, applied mathematics, discrete financial model, exponential growth, growth rate ( r ), annual compounding, financial projection, ( A_n > 10P_0 ), ( A_{n+1} = r A_n ), ( A_5 = 3P_0 )", "For references:\n- Exponential functions in finance\n- Logarithmic transformation in growth modeling\n- Iterative recurrence relations in applied math", "---", "Stay informed on how mathematical models drive sound financial decision-making."]