As for the question,How can I simplify (8x^1/2)/(x^2/3)]^1/3?

[Alex6977]

Hi i have a question i know the answer to but don't understand how it was reached.

[(8x^1/2)/(x^2/3)]^1/3

it comes to 2/x^1/18
but how? I got 8/x^1/18

thx

 

[Greg]

Re: Simplificaiton issues

Hello and welcome! :D

$$\left(\frac{8x^{1/2}}{x^{2/3}}\right)^{1/3}=\left(\frac{8}{x^{1/6}}\right)^{1/3}=\frac{8^{1/3}}{\left(x^{1/6}\right)^{1/3}}=\frac{2}{x^{1/18}}$$

 

[Alex6977]

Re: Simplificaiton issues

I see i tried applying the outside ^1/3 first to eliminate that right off the bat. which would be

8x^1/6 over X^2/9.

but that's still corrrect so far right?

then that equals.

8x^3/18 over x^4/18. but then it starts to look bad and i get lost. maybe you just can't do it that way?

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Oh and thanks

 

[MarkFL]

Your first step is valid, you just need to remember to apply it to the 8 as well, so that you have:

\(\displaystyle \left(\frac{8x^{1/2}}{x^{2/3}}\right)^{1/3}=\frac{2x^{1/6}}{x^{2/9}}\)

Your next step is good too, to convert the rational exponents to a common denominator so they can be subtracted:

\(\displaystyle \left(\frac{8x^{1/2}}{x^{2/3}}\right)^{1/3}=\frac{2x^{3/18}}{x^{4/18}}=\frac{2}{x^{1/18}}\)

 

[Alex6977]

I can't follow that. sorry

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nevermind its displaying right now

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AHHHHH i didn't apply it to the 8. THere we go. Thanks a ton

 

[MarkFL]

I can't follow that. sorry

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nevermind its displaying right now

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AHHHHH i didn't apply it to the 8. THere we go. Thanks a ton

Yeah, I had some missing brackets in my $\LaTeX$ code, and I didn't preview, so you saw a real mess before I got it fixed. :D