Question 1[Maximum number: 3]It is given that log10a=13\log _{10} a=\frac{1}{3}log10a=31, where a>0.Find the value oflog1000a\log _{1000} alog1000a.Mark as masteredShow Answerlog10alog101000\frac{\log _{10} a}{\log _{10} 1000}log101000log10a OR 13log100010\frac{1}{3} \log _{1000} 1031log100010 OR log10001000133\log _{1000} \sqrt[3]{1000^{\frac{1}{3}}}log10003100031 OR 1013=1000x(=(103)x)10^{\frac{1}{3}}=1000^{x}\left(=\left(10^{3}\right)^{x}\right)1031=1000x(=(103)x)log10a3\frac{\log _{10} a}{3}3log10a OR 13log1000100013\frac{1}{3} \log _{1000} 1000^{\frac{1}{3}}31log1000100031 OR log1000100019\log _{1000} 1000^{\frac{1}{9}}log1000100091 OR 3x=133 x=\frac{1}{3}3x=31=19=\frac{1}{9}=91Add to Test