Logarithm simplify 2logbase8ofx^27

[flyingfreedom]

I encountered this question and I can't figure it out.
simplify 2logbase8 of x^27

The answer's available are:
3x
9x
x^3
x^9
The correct answer is x^9
I tried solving but didn't get very far. here's my work:
2logbase8 of x^27
=54logbase8 of x
I don't know where to go from there. I tried turning into exponential form but ended up with this:
8^y=x^54
and that has two variables so I couldn't solve that.
Basically I haven't the slightest idea.
I really need to solve this question
If anybody could help me out that would be great
Thanks,

 

[skeeter]

reason you can't figure it out is because it's not true ...

\(\displaystyle 2\log_8(x^{27}) \neq x^9\)

if x = 1, the left side = 0 and the right side = 1

check the problem statement again ... either you misread/miscopied or the source has some sort of typo.

 

[flyingfreedom]

I misread the question it's actually 2^(logbase8ofx^27)
so then that = 2^(27logbase8 of x)
= 2^(3logbase8 of x^9)
=x^9
What's the principle for turning 2^(3logbase8 of x^9) into x^9
Thakns,

 

[soroban]

Hello, flyingfreedom!

The answer is correct, but those steps are fuzzy . . .

\(\displaystyle 2^{\log_8(x^{27})}\)

\(\displaystyle \text{Then: }\;2^{27\log_8(x)} \;= \;2^{3\log_8(x^9)} \;=\;x^9\)

\(\displaystyle \text{We have: }\;2^{\log_8(x^{27})} \;=\;2^{\log_8(x^{9\cdot3})} \;=\; 2^{3\cdot\log_8(x^9)} \;=\;\left(2^3\right)^{\log_8(x^9)} \;= \;8^{\log_8(x^9)} \;=\; x^9\)

Got it?

 

[flyingfreedom]

yep