CISE 8 Solve the following equations rounded off to three decimal places: \( \begin{array}{lll}\text { (a) } 3^{x}=4 & \text { (b) } 4^{2 x}=5 & \text { (c) } 3^{2 x+1}=12\end{array} \)

To solve the given equations, we can take the logarithm of both sides. **(a) \( 3^{x}=4 \)** Taking the logarithm, we have: \[ x \cdot \log(3) = \log(4) \] Thus, we can find \( x \): \[ x = \frac{\log(4)}{\log(3)} \approx \frac{0.602}{0.477} \approx 1.259 \] **(b) \( 4^{2x}=5 \)** Taking the logarithm gives: \[ 2x \cdot \log(4) = \log(5) \] Solving for \( x \) gives: \[ x = \frac{\log(5)}{2 \cdot \log(4)} \approx \frac{0.699}{2 \cdot 0.602} \approx 0.580 \] **(c) \( 3^{2x+1}=12 \)** Taking the logarithm results in: \[ (2x + 1) \cdot \log(3) = \log(12) \] Expanding this gives: \[ 2x + 1 = \frac{\log(12)}{\log(3)} \] Now let's isolate \( x \): \[ 2x = \frac{\log(12)}{\log(3)} - 1 \] Then, solving for \( x \): \[ x = \frac{1}{2} \left( \frac{\log(12)}{\log(3)} - 1 \right) \approx \frac{1}{2} \left( \frac{1.079}{0.477} - 1 \right) \approx 0.469 \] Final approximations rounded to three decimal places are: - (a) \( x \approx 1.259 \) - (b) \( x \approx 0.580 \) - (c) \( x \approx 0.469 \)