Question

Calculeaza: $\frac{log_{3} 162 }{log_{6}3 } - \frac{log_{3} 54 }{log_{18} 3 }$

Answer 1

$\frac{ log_{3}(162) }{ log_{6}(3) } - \frac{ log_{3}(54) }{ log_{18}(3) } = \frac{ log_{3}(6 \times 27) }{ log_{6}(3) } - \frac{ log_{3}(18 \times 3) }{ log_{18}(3) }$

$= \frac{ log_{3}(6) + log_{3}(27) }{ log_{6}(3) } - \frac{ log_{3}(18) + log_{3}(3) }{ log_{18}(3) }$

$= \frac{ log_{3}(6) + 3}{ log_{6}(3) } - \frac{ log_{3}(18) + 1}{ log_{18}(3) }$

$= log_{3}(6) ( log_{3}(6) + 3) - log_{3}(18) ( log_{3}(18) + 1)$

$= 2 log_{3}(6) + 3 log_{3}(6) - 2 log_{3}(18) - log_{3}(18)$

$= log_{3}( {6}^{2} ) + log_{3}( {6}^{3} ) - log_{3}( {18}^{2} ) - log_{3}(18)$

$= log_{3}(36) + log_{3}(216) - log_{3}(324) - log_{3}(18)$

$= log_{3}( \frac{\frac{36 \times 216}{324} }{18} ) = log_{3}( \frac{ \frac{7776}{324} }{18} ) = log_{3}( \frac{24}{18} ) = log_{3}( \frac{4}{3} )$