Find the value of the expression:
${log}_{10} 64 + {log}_{10} 25 + {log}_{10} 128 - {log}_{10} 5 - 12 {log}_{10} 2$

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Solution

we know that
$l o g_{10} a^{n} = n l o g_{10} a$
$l o g_{10} a b = l o g_{10} a + l o g_{10} b$

So $l o g_{10} 64 + l o g_{10} 25 + l o g_{10} 128 - l o g_{10} 5 - 12 l o g_{10}$
$l o g_{10} 64 + l o g_{10} 25 + l o g_{10} 128 - l o g_{10} 5 - 12 l o g_{10} 2 \Rightarrow l o g_{10} 2^{6} + l o g_{10} 5^{2} + l o g_{10} 2^{7} - l o g_{10} 5 - 12 l o g_{10} 2 \Rightarrow 6 l o g_{10} 2 + 2 l o g_{10} 5 + 7 l o g_{10} 2 - l o g_{10} 5 - 12 l o g_{10} 2 \Rightarrow l o g_{10} 2 + l o g_{10} 5 \Rightarrow l o g_{10} 2 \times 5 = 1$

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