9 Select the correct answer from each drop-down menu. City officials use the given system of equations to estimate the population of two neighboring communities, where \( y \) is the population and \( x \) is the time, in years. \( y=10,000(1.01)^{x} \) \( y=8,000(1.02)^{x} \) Use this system to complete the statement. After about \( \hat{v} \) years, the population of each community will be approximately

Solve the equation by following steps: - step0: Solve for \(x\): \(10000\times 1.01^{x}=8000\times 1.02^{x}\) - step1: Convert the expressions: \(10000\left(\frac{101}{100}\right)^{x}=8000\left(\frac{51}{50}\right)^{x}\) - step2: Multiply both sides: \(\frac{1}{10000}\times 10000\left(\frac{101}{100}\right)^{x}=\frac{1}{10000}\times 8000\left(\frac{51}{50}\right)^{x}\) - step3: Calculate: \(\left(\frac{101}{100}\right)^{x}=\frac{4}{5}\left(\frac{51}{50}\right)^{x}\) - step4: Divide both sides: \(\frac{\left(\frac{101}{100}\right)^{x}}{\left(\frac{51}{50}\right)^{x}}=\frac{4}{5}\) - step5: Evaluate: \(\left(\frac{101}{102}\right)^{x}=\frac{4}{5}\) - step6: Take the logarithm of both sides: \(\log_{\frac{101}{102}}{\left(\left(\frac{101}{102}\right)^{x}\right)}=\log_{\frac{101}{102}}{\left(\frac{4}{5}\right)}\) - step7: Evaluate the logarithm: \(x=\log_{\frac{101}{102}}{\left(\frac{4}{5}\right)}\) - step8: Simplify: \(x=2\log_{\frac{101}{102}}{\left(2\right)}-\log_{\frac{101}{102}}{\left(5\right)}\) To find the time \( \hat{v} \) in years when the populations of the two communities will be approximately equal, we start with the given equations: 1. \( y = 10,000(1.01)^{x} \) (for the first community) 2. \( y = 8,000(1.02)^{x} \) (for the second community) We set these two equations equal to each other to find \( x \): \[ 10,000(1.01)^{x} = 8,000(1.02)^{x} \] Next, we can simplify this equation: \[ \frac{10,000}{8,000} = \frac{(1.02)^{x}}{(1.01)^{x}} \] This simplifies to: \[ 1.25 = \left(\frac{1.02}{1.01}\right)^{x} \] Taking the logarithm of both sides gives us: \[ \log(1.25) = x \cdot \log\left(\frac{1.02}{1.01}\right) \] Solving for \( x \): \[ x = \frac{\log(1.25)}{\log\left(\frac{1.02}{1.01}\right)} \] Using the change of base formula, we can express this as: \[ x = 2\log_{\frac{101}{102}}(2) - \log_{\frac{101}{102}}(5) \] Now, we can calculate the approximate value of \( x \) to find \( \hat{v} \). Let's calculate \( \hat{v} \). Calculate the value by following steps: - step0: Calculate: \(\frac{2\log_{10}{\left(2\right)}}{\left(\log_{10}{\left(101\right)}-\log_{10}{\left(102\right)}\right)}-\frac{\log_{10}{\left(5\right)}}{\left(\log_{10}{\left(101\right)}-\log_{10}{\left(102\right)}\right)}\) - step1: Remove the parentheses: \(\frac{2\log_{10}{\left(2\right)}}{\log_{10}{\left(101\right)}-\log_{10}{\left(102\right)}}-\frac{\log_{10}{\left(5\right)}}{\log_{10}{\left(101\right)}-\log_{10}{\left(102\right)}}\) - step2: Use the logarithm product rule: \(\frac{2\log_{10}{\left(2\right)}}{\log_{10}{\left(\frac{101}{102}\right)}}-\frac{\log_{10}{\left(5\right)}}{\log_{10}{\left(101\right)}-\log_{10}{\left(102\right)}}\) - step3: Use the logarithm product rule: \(\frac{2\log_{10}{\left(2\right)}}{\log_{10}{\left(\frac{101}{102}\right)}}-\frac{\log_{10}{\left(5\right)}}{\log_{10}{\left(\frac{101}{102}\right)}}\) - step4: Use the logarithm base change rule: \(2\log_{\frac{101}{102}}{\left(2\right)}-\frac{\log_{10}{\left(5\right)}}{\log_{10}{\left(\frac{101}{102}\right)}}\) - step5: Use the logarithm base change rule: \(2\log_{\frac{101}{102}}{\left(2\right)}-\log_{\frac{101}{102}}{\left(5\right)}\) After calculating, we find that \( \hat{v} \approx 22.65 \) years. Thus, we can complete the statement: After about \( \hat{v} \) years, the population of each community will be approximately equal. To summarize: - \( \hat{v} \approx 22.65 \) years. This means that after approximately 23 years, the populations of the two communities will be roughly the same.