Find the value of ${log}_{10} 72$ using log table

0.901 + 0.909

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0.903 + 0.954

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1.890

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2.104

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Solution

The correct option is C $0.903 + 0.954$
${log}_{10} 72 = {log}_{10} (2^{3} {.3}^{2})$
$= {log}_{10} 2^{3} + {log}_{10} 3^{2}$
$= 3. {log}_{10} 2 + 2. {log}_{10} 3$
$= 3 (0.301) + 2 (0.4771)$
$= 0.903 + 0.954$

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