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

$rArr 6^1=t...........................[because, (lambda)]$ .

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

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

$:. 5.5x=64$ .

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

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