Algebra -exponents

[CSmith]

is this correct?
1.) 32 2/5


(32 1/5) ^2


=(5 SQUARE ROOT 32)2

=5(SQUAREROOT 2X2X2X2X2)2
=(5 SQUARE ROOT )2

=5 SQUARE ROOT 2 X 5 SQUARE ROOT 2
=25 (2 SQUARE ROOT 2)
=50 SQUARE ROOT 2

 

[SuperSonic4]

is this correct?
1.) $32^{2/5}$


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


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

=5(SQUAREROOT 2X2X2X2X2)2
$= (\sqrt[5]{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2})^2$

Notation aside you're correct up to here

=(5 SQUARE ROOT )2

=5 SQUARE ROOT 2 X 5 SQUARE ROOT 2
=25 (2 SQUARE ROOT 2)
=50 SQUARE ROOT 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 $2^2$

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.

 

[Jameson]

I don't know how you were taught to approach these, but the short way to remember a fraction as an exponent is "power over root". The top number of the fraction is a power, so you should multiply the number together that many times. The bottom number is a root, so you need to find a number that multiplied together that many times gives you the original number. You can do it in either order, but you do one at a time. You have 32 to begin with so it's logical to take the root first because that will be a smaller number than 32. If you apply the power first then the result will be quite large and less easy to work with.

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