How do you solve log _ 6 ( log _ 2 (5.5x)) = 1log6(log2(5.5x))=1?

2 Answers
Mar 4, 2018

x=128/11=11.bar(63)x=12811=11.¯¯¯¯63

Explanation:

We begin by raising both sides as a power of 66:

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)

Mar 4, 2018

x=128/11~~11.64

Explanation:

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