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\[ b = 4 \] - Portal da AcÃºstica

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 oxed{egin{cases} T = egin{pmatrix} a & b \ c & -a \end{pmatrix}, & a^2 + bc = 1 \end{cases}}

📌 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 \( A_n > 10P_0 \).

📌 Solution:** From the recurrence \( A_{n} = r^n P_0 \), and given \( A_5 = 3P_0 \), we have:

📌 r^5 P_0 = 3P_0 \Rightarrow r^5 = 3 \Rightarrow r = 3^{1/5}

📌 The annual growth rate is \( r = 3^{1/5} \). Now find the smallest integer \( n \) such that \( A_n > 10P_0 \):

📌 r^n > 10 \Rightarrow (3^{1/5})^n > 10 \Rightarrow 3^{n/5} > 10 \Rightarrow rac{n}{5} > \log_3 10

📌 Using \( \log_3 10 = rac{\ln 10}{\ln 3} pprox rac{2.3026}{1.0986} pprox 2.095 \), we get:

📌 rac{n}{5} > 2.095 \Rightarrow n > 10.475 \Rightarrow n \geq 11

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