How do you solve #log _ 6 ( log _ 2 (5.5x)) = 1#?

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Answer 1 · 2018-03-04T10:31:27

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

We begin by raising both sides as a power of #6#:

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

#log_2(5.5x)=6#

Then we raise both sides as powers of #2#:

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

#5.5x=64#

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

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

Answer 2 · 2018-03-04T10:32:58

# x=128/11~~11.64#

Recall that #log_ba=m iff b^m=a..........(lambda)#.

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

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

#rArr 6^1=t...........................[because, (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#