[-(1+a)] If \log_{10}{7}=a, then \log_{10}({\frac{1}{70}}) is equal to:

For Competitive Exams

If $\log_{10}{7}=a, then \log_{10}({\frac{1}{70}}) $is equal to:

-(1+a)

$(1+a)^{-1}$

$\frac{a}{10}$

$\frac{1}{10a}$

Explanation:

$\log_{10}({\dfrac{1}{70}})$= $\log_{10}{1}- \log_{10}{70}$
=>$- \log_{10}{7\times 10}$
=>-(a+1)

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