81. STOCK The price per share of a coffee chain's stock was \( \$ 0.93 \) in a month during its first year of trading. During its fifth year of trading, the price per share of stock was \( \$ 3.52 \) during the same month. (Example 9) a. Write a continuous exponential equation to model the price of stock \( P \) as a function of year of trading \( t \). Round \( k \) to the nearest ten-thousandth. b. Use the model you found in part a to predict the price of the stock during the ninth year of trading.

Given: - The price per share of a coffee chain's stock was $0.93 in the first year of trading. - The price per share of stock was $3.52 in the fifth year of trading. We can model the price of stock \( P \) as a function of the year of trading \( t \) using a continuous exponential equation. The general form of an exponential equation is: \[ P = P_0 \times e^{kt} \] where: - \( P_0 \) is the initial price (price in the first year), - \( k \) is the growth rate, - \( t \) is the year of trading. a. To find the growth rate \( k \), we can use the given information: - In the first year, \( P = 0.93 \) and \( t = 1 \). - In the fifth year, \( P = 3.52 \) and \( t = 5 \). We can set up two equations using the exponential equation: \[ 0.93 = 0.93 \times e^{k \times 1} \] \[ 3.52 = 0.93 \times e^{k \times 5} \] Now, we can solve these equations to find the value of \( k \). b. Once we have the value of \( k \), we can use the model to predict the price of the stock during the ninth year of trading by substituting \( t = 9 \) into the exponential equation. Let's solve the equations to find the value of \( k \) and then use the model to predict the price of the stock during the ninth year of trading. Solve the equation by following steps: - step0: Solve for \(k\): \(0.93=0.93e^{k\times 1}\) - step1: Multiply: \(0.93=0.93e^{k}\) - step2: Swap the sides: \(0.93e^{k}=0.93\) - step3: Divide both sides: \(\frac{0.93e^{k}}{0.93}=\frac{0.93}{0.93}\) - step4: Divide the numbers: \(e^{k}=1\) - step5: Write in exponential form: \(e^{k}=e^{0}\) - step6: Set the exponents equal: \(k=0\) Solve the equation \( 3.52 = 0.93 * e^(k * 5) \). Solve the equation by following steps: - step0: Solve for \(k\): \(3.52=0.93e^{k\times 5}\) - step1: Reorder the terms: \(3.52=0.93e^{5k}\) - step2: Swap the sides: \(0.93e^{5k}=3.52\) - step3: Divide both sides: \(\frac{0.93e^{5k}}{0.93}=\frac{3.52}{0.93}\) - step4: Divide the numbers: \(e^{5k}=\frac{352}{93}\) - step5: Take the logarithm of both sides: \(\ln{\left(e^{5k}\right)}=\ln{\left(\frac{352}{93}\right)}\) - step6: Evaluate the logarithm: \(5k=\ln{\left(\frac{352}{93}\right)}\) - step7: Divide both sides: \(\frac{5k}{5}=\frac{\ln{\left(\frac{352}{93}\right)}}{5}\) - step8: Divide the numbers: \(k=\frac{\ln{\left(\frac{352}{93}\right)}}{5}\) - step9: Simplify: \(k=\frac{\ln{\left(352\right)}-\ln{\left(93\right)}}{5}\) Calculate \( k=\frac{\ln{(352)}-\ln{(93)}}{5} \). Calculate the value by following steps: - step0: Calculate: \(\frac{\ln{\left(352\right)}-\ln{\left(93\right)}}{5}\) The growth rate \( k \) is approximately 0.266206. Now, we can use the model to predict the price of the stock during the ninth year of trading by substituting \( t = 9 \) into the exponential equation: \[ P = 0.93 \times e^{0.266206 \times 9} \] Let's calculate the predicted price of the stock during the ninth year of trading. Calculate the value by following steps: - step0: Calculate: \(0.93e^{0.266206\times 9}\) - step1: Multiply the numbers: \(0.93e^{2.395854}\) - step2: Convert the expressions: \(0.93e^{\frac{1197927}{500000}}\) - step3: Convert the expressions: \(\frac{93}{100}e^{\frac{1197927}{500000}}\) - step4: Multiply the numbers: \(\frac{93e^{\frac{1197927}{500000}}}{100}\) The predicted price of the stock during the ninth year of trading is approximately $10.21.