The value of $l o g 0.008........$

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Solution

$l o g_{10} 0.008$
$\Rightarrow l o g_{10} 8 \times 10^{- 3}$ [ log ab = log a+ log b]
$\Rightarrow l o g_{10} 8 + l o g_{10} 10^{- 3}$
$\Rightarrow l o g_{10} 2^{3} + (- 3) l o g_{10} 10$
$\Rightarrow 3 l o g_{10} 2 + (- 3)$ [ log ab = b log a]
$\Rightarrow 3 (0.301) - 3$
$\Rightarrow - 2.097$

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