log-6-2-a-log-6-9-

Question Number 197249 by sciencestudentW last updated on 11/Sep/23

lo \underset{6}{g 2} = a

lo \underset{6}{g 9} = ?

Answered by AST last updated on 11/Sep/23

\frac{l o g 2}{l o g 2 + l o g 3} = a \Rightarrow l o g 2 (1 - a) = a l o g 3

\frac{2 l o g 3}{l o g 2 + l o g 3} = \frac{2}{\frac{l o g 2}{l o g 3} + 1} = \frac{2}{\frac{a}{1 - a} + 1} = \frac{\frac{2}{1}}{\frac{1}{1 - a}} = 2 - 2 a

Answered by Rasheed.Sindhi last updated on 13/Sep/23

\log_{6} 2 = a; \log_{6} 9 = ?

\log_{6} 2 = a \Rightarrow \log_{6} (\frac{6}{3}) = a

\log_{6} 6 - \log_{6} 3 = a

1 - \log_{6} 3 = a

\log_{6} 3 = 1 - a

▸ \log_{6} 9 = \log_{6} 3^{2} = {2 \log}_{6} 3 = 2 (1 - a)

Answered by Rasheed.Sindhi last updated on 13/Sep/23

\log_{6} 2 = a; \log_{6} 9 = ?

a = \log_{6} 2

= \frac{1}{2} \log_{6} 4 = \frac{1}{2} \log_{6} (\frac{36}{9})

= \frac{1}{2} {\log_{6} 36 - \log_{6} 9}

= \frac{1}{2} {2 - \log_{6} 9} = a

= 1 - \frac{\log_{6} 9}{2} = a

\log_{6} 9 = 2 (1 - a)