Question on indices

[Guest]

I am trying to evaluate [9^(1/3)*27^(-1/2)] ÷ [3^(-1/6)*3^(-2/3)] and have got stuck. Can anyone help? My working is below:

Multiply by 3^(1/6)*3^(2/3) to remove negative indices from denominator...

9^(1/3)*3^(1/6)*3^(2/3)*27^(-1/2)

Simplify expression by adding together the 3's...

3^(1/6)*3^(4/6) = 3^(5/6)

This leaves me with:
9^(1/3)*3^(2/3)*27^(-1/2)

From here I can't multiply the bases because they all have different indices, nor can I add the indices because they all have different bases. How do I get from here to the final answer of 1?

:?

 

[pka]

It would help if you first wrote:
\(\displaystyle \L
\begin{array}{l}
{\rm{9}}^{{\rm{1/3}}} = \left( {3^2 } \right)^{1/3} = 3^{2/3} \\
27^{ - 1/2} = 3^{ - 3/2} \\
\end{array}\)

 

[stapel]

I am reading your exercise to be as follows:


. . . . .\(\displaystyle \large{\frac{9^{\frac{1}{3}}\,27^{-\frac{1}{2}}}{3^{-\frac{1}{6}}\,3^{-\frac{2}{3}}}}\)


I would suggest converting everything to powers of 3:


. . . . .\(\displaystyle \large{\frac{(3^2)^{\frac{1}{3}}\,(3^3)^{-\frac{1}{2}}}{3^{-\frac{1}{6}}\,3^{-\frac{2}{3}}}}\)


. . . . .\(\displaystyle \large{\frac{3^{\frac{2}{3}}\,3^{-\frac{3}{2}}}{3^{-\frac{1}{6}\,-\,\frac{4}{6}}}}\)


. . . . .\(\displaystyle \large{\frac{3^{\frac{4}{6}\,-\,\frac{9}{6}}}{3^{-\frac{5}{6}}}\)


Continue with the simplification.

Eliz.

 

[Gene]

If you were evaluating
2+2^(-1) would you multiply by 2^1 and get 2*2^1+1???
That only works if it is a fraction and you are putting it over a common denominator.
You do want to multiply the RHS by 2^1/2^1.
2+2^(-1)*2^1/2^2^1 =
2+2^0/2^1 =
2+1/2^1
Or you can just move the negative exponents across the line and change the sign. It doesn't affect the first 2 (or in this problem it would become
[9^(1/3)/27^(1/2)] ÷ 1/[3^(1/6)*3^(2/3)]
But it isn't really necessary. You can just leave the exponents negative but you want the base to be the same so you can add them.
[9^(1/3)*27^(-1/2)] ÷ [3^(-1/6)*3^(-2/3)] =

[(3^2)^(1/3)*(3^3)^(-1/2)] ÷ [3^(-1/6)*3^(-2/3)] =

[3^(2/3)*3^(-3/2)] ÷ [3^(-1/6)*3^(-2/3)] =

3^(4/6-9/6) + 3^(-1/6-4/6) =
3^(-5/6)+3^(-5/6)

After all that if you are expecting an answer of 1 you have a typo. You probably want
[9^(1/3)*27^(-1/2)] / [3^(-1/6)*3^(-2/3)]

Most of what I did applies either way.

 

[stapel]

As you can see from the above thread, ben654, there is more than one way to approach this exercise. Please reply, showing your steps, if you have difficulty following through to the solution.

Thank you.

Eliz.