log-x-16-log-16-x-log-512-x-log-x-512-x-

Question Number 94456 by Abdulrahman last updated on 18/May/20

\log_{x} 16 + \log_{16} x = \log_{512} x + \log_{x} 512

x = ?

Answered by john santu last updated on 19/May/20

4 (\log_{x} (2)) + \frac{1}{4} (\log_{2} (x)) =

9 (\log_{x} (2)) + \frac{1}{9} (\log_{2} (x))

\Rightarrow 5 (\log_{x} (2)) = \frac{5}{36} (\log_{2} (x))

36 (\log_{x} (2)) = \frac{1}{\log_{x} (2)}

{\begin{cases} \log_{x} (2) = \frac{1}{6} \Rightarrow \frac{1}{\log_{2} (x)} = \frac{1}{6}; x = 64 \\ \log_{x} (2) = - \frac{1}{6} \Rightarrow \frac{1}{\log_{2} (x)} = - \frac{1}{6} \end{cases}

x = \frac{1}{64}

Commented by Abdulrahman last updated on 19/May/20

very good

but i didnt know in two final lines

Commented by i jagooll last updated on 19/May/20

nice solution

Commented by i jagooll last updated on 19/May/20

why sir ?

{(\log_{x} (2))}^{2} = \frac{1}{36} . l e t \log_{x} (2) = p

p^{2} - \frac{1}{36} = 0

(p - \frac{1}{6}) (p + \frac{1}{6}) = 0 s i r

Commented by Abdulrahman last updated on 19/May/20

thanks alot now i got it