Question 1[Maximum number: 2]It is given that log10a=13\log _{10} a=\frac{1}{3}log10a=31, where a>0.Find the value oflog10(1a)\quad \log _{10}\left(\frac{1}{a}\right)log10(a1);Mark as masteredShow Answerlog101−log10a\log _{10} 1-\log _{10} alog101−log10a OR log10a−1=−log10a\log _{10} a^{-1}=-\log _{10} alog10a−1=−log10a OR log1010−13\log _{10} 10^{-\frac{1}{3}}log1010−31 OR 10x=1101310^{x}=\frac{1}{10^{\frac{1}{3}}}10x=10311=−13=-\frac{1}{3}=−31Add to Test