91Ó°ÊÓ

Evaluate lg \(0.001\) Let \(x=\lg 0.001=\log _{10} 0.001 \quad\) then \(10^{x}=0.001\) i.e. \(\quad 10^{x}=10^{-3} \quad\) from which, \(x=-3\) Hence, \(\quad\) lg \(0.001=-3\) (which may be checked using a calculator)

Simplify: \(\log 64-\log 128+\log 32\) \(64=2^{6}, 128=2^{7}\) and \(32=2^{5}\) Hence, \(\log 64-\log 128+\log 32\) $$ \begin{aligned} &=\log 2^{6}-\log 2^{7}+\log 2^{5} \\ &=6 \log 2-7 \log 2+5 \log 2 \end{aligned} $$ by the third law of logarithms \(=4 \log 2\)