Solution 3.3:2h

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The argument in the logarithm can be rewritten as \displaystyle \frac{1}{10^{2}}=10^{-2} and then the log law \displaystyle \lg a^{b}=b\lg a gives the rest:


\displaystyle \lg \frac{1}{10^{2}}=\lg 10^{-2}=\left( -2 \right)\centerdot \lg 10=\left( -2 \right)\centerdot 1=-2