For some positive real number {\x}, if \log _{\sqrt{3}}(x)+\frac{\log _x(25)}{\log _x(0.008)}=\frac{16}{3}, then the value of \log _3 (3 x^2) is

CAT 2023 Slot 2 Solved Paper

Question : 50 of 66

Marks: +1, -0

For some positive real number

{\x}

, if

\log _{{√{3}}}(x)+\;{\log _x(25)}/{\log _x(0.008)}=\;{16}/{3}

, then the value of

\log _3 (3 x^2)

Solution:

It is given that

\log _{{√{3}}}(x)+\;{\log _x(25)}/{\log _z(0.008)}=\;{16}/{3}

, which can be written as:

\;⇒2 \log _3 x+\log _{0.008} 25=\;{16}/{3}

\;⇒2 \log _3 x+\log _{\;{8}/{1000}} 25=\;{16}/{3}

\;⇒2 \log _3 x+\log _{\;{1}/{125}} 25=\;{16}/{3}

\;⇒2 \log _3 x+\log _{5-3}(5)^2=\;{16}/{3}

\;⇒2 \log _3 x-\;{2}/{3}=\;{16}/{3}

\;⇒2 \log _3 x=\;{16}/{3}+\;{2}/{3}

\;⇒2 \log _3 x=6

\;⇒\log _3 x^2=6⇒x^2=3^6

Hence,

\log _3 (3 ⋅ x^2) =\log _3 (3 ⋅ 3^6) =\log _3 3^7=7

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