Find the value of the expression:
$l o g_{10} 64 + l o g_{10} 25 + l o g_{10} 128 - l o g_{10} 5 - 12 l o g_{10} 2$ .

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Solution

The correct option is B 1
$l o g_{10} 64 + l o g_{10} 25 + l o g_{10} 128 - l o g_{10} 5 - 12 l o g_{10} 2$
$= l o g_{10} [\frac{(64) (25) (128)}{(5) (2^{12})}]$
$= l o g_{10} [\frac{204800}{20480}] = l o g_{10} 10 = 1$

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Similar questions

{log}_{10} 5 \cdot {log}_{10} 20 + ({log}_{10} 2)^{2}

{log}_{10} 5. {log}_{10} 20 + ({log}_{10} 2)^{2}

{log}_{10} 5

{log}_{10} 2 = 0.301

a = 1 + l o g_{10} 2 - l o g_{10} 5, b = 2 l o g_{10} 3

c = l o g_{10} m - l o g_{10} 5,

{log}_{10} 4 = 0.6021

{log}_{10} 5 = 0.6990

{log}_{10} 1600

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