(3) Evaluate, without using tables, \( \log _{10} 1.44-\log _{10} 90+\log _{10} 0.062 \)

To evaluate \( \log_{10} 1.44 - \log_{10} 90 + \log_{10} 0.062 \), we can use properties of logarithms. According to the subtraction property, \( \log_a b - \log_a c = \log_a \left(\frac{b}{c}\right) \). First, combine the first two logs: \[ \log_{10} 1.44 - \log_{10} 90 = \log_{10} \left(\frac{1.44}{90}\right) \] Calculating \( \frac{1.44}{90} = 0.016. \) Now we can rewrite the expression as: \[ \log_{10} 0.016 + \log_{10} 0.062 \] Using the addition property of logarithms, this becomes: \[ \log_{10} (0.016 \times 0.062) \] Calculating \( 0.016 \times 0.062 = 0.000992. \) Thus, we have: \[ \log_{10} 0.000992 \] Now, we simplify \( 0.000992 \) as \( 992 \times 10^{-6} \). The logarithm can be expressed as: \[ \log_{10} 992 - 6 \] To find \( \log_{10} 992 \), we estimate it close to \( 3 \) because \( 10^3 = 1000 \) and \( 10^2 = 100 \). So: \[ \log_{10} 992 \approx 3 \] Thus, \[ \log_{10} 0.000992 \approx 3 - 6 = -3 \] Therefore, the final answer is: \[ \log_{10} 1.44 - \log_{10} 90 + \log_{10} 0.062 \approx -3 \]