- Thread starter
- #1

- Thread starter CSmith
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- Thread starter
- #1

- Mar 1, 2012

- 249

Notation aside you're correct up to hereis this correct?

1.) $32^{2/5}$

$(32^{1/5}) ^2$

=(5 SQUARE ROOT 32)2

$(\sqrt[5]{32})^2$

=5(SQUAREROOT 2X2X2X2X2)2

$= (\sqrt[5]{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2})^2$

You've lost me here though. You have $(\sqrt[5]{2^5})^2$ which is the same as $((2^5)^{1/5})^2$ which cancels down to $2^2$=(5 SQUARE ROOT )2

=5 SQUARE ROOT 2 X 5 SQUARE ROOT 2

=25 (2 SQUARE ROOT 2)

=50 SQUARE ROOT 2

A simpler way IMO is to note that $32 = 2^5$ (as you did). That leaves you with $(2^5)^{2/5} = 2^{5 \cdot 2/5} = 2^2$

Your notation is very tricky to follow. Please use Latex or "^" together with brackets next time to make it easer to follow.

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- #3

- Jan 26, 2012

- 4,041

\(\displaystyle 32^{\frac{1}{5}}=2\) because $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$.

So now we've applied the root and got 2. Now we apply the power and get $2^2=4$ and we're done. Again you can choose the order in which you calculate this so take a second to consider both options and choose the one that has the easier numbers to work with.