If log _{10} 7 = a,then log _{10} left( frac{ | vedclass

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(A) Given $\log _{10} 7 = a$.Using the property $\log \left( \frac{x}{y} \right) = \log x - \log y$,we have:$\log _{10} \left( \frac{1}{70} \right) =

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$-(1+a)$

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(A) Given $\log _{10} 7 = a$.Using the property $\log \left( \frac{x}{y} ight) = \log x - \log y$,we have:$\log _{10} \left( \frac{1}{70} ight) = \log _{10} 1 - \log _{10} 70$.Since $\log _{10} 1 = 0$ and $\log _{10} (70) = \log _{10} (7 	imes 10)$:$= 0 - \log _{10} (7 	imes 10)$.Using the property $\log (xy) = \log x + \log y$:$= -[\log _{10} 7 + \log _{10} 10]$.Since $\log _{10} 7 = a$ and $\log _{10} 10 = 1$:$= -(a + 1) = -(1 + a)$.

If $\log _{10} 7 = a$,then $\log _{10} \left( \frac{1}{70} \right)$ is equal to

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