Log₁₂27=log₁₂3³=3log₁₂3
3log₁₂3=a
log₁₂3=a/3
1/log₃12=a/3
log₃3*4=3/a
log₃3+log₃2²=3/a
1+2log₃2=3/a
2log₃2=3/a-1
log₃2=(3-a)/2a
Теперь выразим log₆16 через log₃2.
имеем log₆16=log₆2⁴=4log₆2=4/log₂6=4/log₂2*3=4/(log₂2+log₂3)= 4/(1+log₂3)
log₂3=1/log₃2.=2a/(3-a)
Тогда (1+log₂3) =1+ (2a/(3-a))=(3-а+2а)/(3-а)=(а+3)/(3-а).
log₆16= 4/(1+log₂3)=4*(3-а)/(а+3)