Kako rešite log_6 (log _ 2 (5.5x)) = 1?

Odgovor:

# x = 128/11 = 11.bar (63) #

Pojasnilo:

Začnemo z dvigom obeh strani kot moči #6#:

# cancel6 ^ (prekliči (log_6) (log_2 (5.5x))) = 6 ^ 1 #

# log_2 (5.5x) = 6 #

Potem dvignemo obe strani kot moči #2#:

# cancel2 ^ (prekliči (log_2) (5.5x)) = 2 ^ 6 #

# 5.5x = 64 #

# (cancel5.5x) /cancel5.5=64/5.5#

# x = 128/11 = 11.bar (63) #

Odgovor:

# x = 128/11 ~~ 11.64 #

Pojasnilo:

Spomnimo se tega # log_ba = m iff b ^ m = a ………. (lambda) #.

Pusti, # log_2 (5.5x) = t #.

Potem, # log_6 (log_2 (5.5x)) = 1 rArr log_6 (t) = 1 #.

#rArr 6 ^ 1 = t ……………………… ker, (lambda) #.

#rArr t = log_2 (5.5x) = 6 #.

#:. "By" (lambda), 2 ^ 6 = 5.5x #.

#:. 5,5x = 64 #.

#rArr x = 64 / 5.5 = 128/11 ~~ 11.64 #