Cách giải logarit của Cảnh bằng cách đưa về cùng cơ số

Bài 1: log₂log₂X=log₃log₃x

Bài 2: log₃log₂X=log₂log₃x

Bài 3: log₂log₃x+ log₃log₂X= log₃log₃x

Bài 4: log₂log₃log₄x=log₄log₃log₂x

Giải:

Bài 1: log₂log₂X=log₃log₃x đk (x>0,x≠1)

ó log₂3.log₃log₂x= log₃log₃x (log_ab=log_ac.log_cb)

ó log₂3.log₃(log₂3.log₃x)=log₃log₃x (log_ab.c=log_ab+log_ac)

ó log₂3.(log₃log₂3+ log₃log₃x) =log₃log₃x

ó log₂3.log₃log₂3+ log₂3. log₃log₃x =log₃log₃x

ó log₂3.log₃log₂3= log₃log₃x- log₂3. log₃log₃x

ó log₂3.log₃log₂3= log₃log₃x(1- log₂3)

ó log₂3.log₃log₂3= log₃log₃x(log₂) (1=log₂2, log_ab-log_ac=log_a)

ó log₂log₂3= log₃log₃x(log₂) (log_ab=log_ac.log_cb)

ó= log₃log₃x

ó log₂3=log₃log₃x

ó log₃x=

ó x=

Bài 2: log₃log₂X=log₂log₃x

ó log₃2.log₂log₂X=log₂log₃x (log_ab=log_ac.log_cb)

ólog₃2[log₂(log₂3.log₃X) ]=log₂log₃x

ólog₃2[log₂log₂3+ log₂log₃x]= log₂log₃x

ólog₃2.log₂log₂3+ log₃2. log₂log₃x = log₂log₃x

ó log₃2.log₂log₂3= log₂log₃x- log₃2. log₂log₃x

ó log₃2.log₂log₂3= log₂log₃x(1- log₃2)

ó log₃log₂3= log₂log₃x()

ó = log₂log₃x

= log₂log₃x

ó log₃x=

ó x=

bài 3: log₂log₃x+ log₃log₂X= log₃log₃x

ó log₂3.log₃log₃x+log₃(log₂3.log₃x)=log₃log₃x

ó log₂3.log₃log₃x+ log₃log₂3+ log₃log₃x=log₃log₃x

ó log₂3.log₃log₃x+ log₃log₂3=0

ó log₂3.log₃log₃x= - log₃log₂3

ó log₂3.log₃log₃x= log₃

ó log₂3.log₃log₃x= log₃log₃2

ó log₃log₃x==

ó log₃log₃x=

ó log₃x=

ó x=

Bài 4: log₂log₃log₄x=log₄log₃log₂x

ó log₂log₃(.log₂x)=log₂log₃log₂x

ó log₂log₃+ log₂log₃log₂x=log₂log₃log₂x

ó log₂log₃= -log₂log₃log₂x

ó -2 log₂log₃= log₂log₃log₂x

ó log₂= log₂log₃log₂x

ó = log₃log₂x

ó log₂x=

ó x=

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