Question

Дано $log _{12} 27=a$ Найти $log _{6} 16$

Answer 1

 $log_{12}3^3=a\ log_{12}3= frac{a}{3}\ log_{6}16 = frac{log_{12}16}{log_{12}6}=frac{4log_{12}2}{log_{12}3 + log_{12}2}\ \ log_{3}(4*3)=log_{3}4+1=frac{3}{a}\ log_{3}4=frac{3-a}{a}\ frac{log_{12}4}{log_{12}3}=frac{3-a}{a}\ log_{12}2=frac{3-a}{6}\ \ log_{6}16=frac{-4*frac{(a-3)}{6}}{frac{a+3}{6}}=frac{12-4a}{a+3}$

Answer 2

log₁₂27=3log₁₂3=3log₂3/log₂12=3log₂3/(log₂4+log₂3)=3log₂3/(2+log₂3)=a
3log₂3=2a+alog₂3⇒3log₂3-alog₂3=2a⇒log₂3(3-a)=2a⇒log₂3=2a/(3-a)
log₆16=4log₆2=4log₂2/log₂6=4/(log₂2+log₂3)=4/(1+log₂3)
log₆16=4/(1+2a/(3-a))=4(3-a)/(3-a+2a)=4(3-a)/(3+a)