Find \( x \). \( \begin{array}{ll}\text { (A) } \log x=2.0423 & \text { (B) } \log x=-1.5163 \\ \text { (C) } \ln x=3.0568 & \text { (D) } \ln x=-1.7996\end{array} \)

To find \( x \) for each of the given equations, we will solve for \( x \) using the properties of logarithms. 1. **For (A)**: \( \log x = 2.0423 \) We can rewrite this in exponential form: \[ x = 10^{2.0423} \] Calculating this gives: \[ x \approx 10^{2.0423} \approx 110.4 \] 2. **For (B)**: \( \log x = -1.5163 \) Again, we rewrite this in exponential form: \[ x = 10^{-1.5163} \] Calculating this gives: \[ x \approx 10^{-1.5163} \approx 0.031 \] 3. **For (C)**: \( \ln x = 3.0568 \) We can rewrite this in exponential form: \[ x = e^{3.0568} \] Calculating this gives: \[ x \approx e^{3.0568} \approx 21.2 \] 4. **For (D)**: \( \ln x = -1.7996 \) We rewrite this in exponential form: \[ x = e^{-1.7996} \] Calculating this gives: \[ x \approx e^{-1.7996} \approx 0.165 \] In summary, the values of \( x \) for each option are approximately: - (A) \( x \approx 110.4 \) - (B) \( x \approx 0.031 \) - (C) \( x \approx 21.2 \) - (D) \( x \approx 0.165 \)