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I have an indices related question I've been trying to figure out tonight.

Basically I have the following question:

$$\sqrt{\frac{\left(25z\right)^6}{4z^{-3}}}\frac{2z^{-\frac{1}{2}}}{\left(\left(5z\right)^{-2}\right)^{-3}}$$

I know the answer is $\frac 1{z^2}$

But I have no idea how to get there.

Anyone that can give me some tips?

Feyre
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Johnny Mccrum
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  • How to format your math. –  Sep 08 '15 at 21:16
  • What have you done so far? For instance, how far can you simplify the left part of the equation? People will be more inclined to help if you show your attempts. – Feyre Sep 08 '15 at 21:18
  • they're all on paper, but my attempts were ridiculous, I tried to multiply out the first bracket (25 to the power of 6) and was trying to find the square root of 244140621. I know there's a z3 somewhere too. – Johnny Mccrum Sep 08 '15 at 21:20
  • It's often helpful to only simplify in the exponents. For example, you can write $((5z)^{-2})^{-3} = 5^{6}z^{6}$, but then leave that as it is. This can make it a lot more manageable to write and perform cancellations later. – Jesse Sep 08 '15 at 21:36
  • That's kinda what I've been doing, for the first section I have something like 150z6/4z-3 right now. – Johnny Mccrum Sep 08 '15 at 21:40

2 Answers2

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The basic rules of exponents are $a^ba^c =a^{b+c} $, $a^{-b} =\frac1{a^b} $, and $(a^b)^c =a^{bc} $.

Therefore,

$\begin{array}\\ \sqrt{\frac{\left(25z\right)^6}{4z^{-3}}}\frac{2z^{-\frac{1}{2}}}{\left(\left(5z\right)^{-2}\right)^{-3}} &=\left(\frac{\left(25z\right)^6}{4z^{-3}}\right)^{1/2}\frac{2z^{-\frac{1}{2}}}{\left(5^{-2}z^{-2}\right)^{-3}} \\ &=\left(\frac{25^6z^6}{4z^{-3}}\right)^{1/2}\frac{2z^{-\frac{1}{2}}}{\left(5^{-2}z^{-2}\right)^{-3}} \\ &=\frac{(25^6z^6)^{1/2}}{(4z^{-3})^{1/2}}\frac{2z^{-\frac{1}{2}}}{5^{6}z^{6}} \\ &=\frac{25^3z^3}{2z^{-3/2}}\frac{2z^{-\frac{1}{2}}}{5^{6}z^{6}} \\ &=5^6z^3z^{3/2}\frac{z^{-\frac{1}{2}}}{5^{6}z^{6}} \\ &=z^{3+3/2-1/2-6}\\ &=z^{-2}\\ &=\frac{1}{z^{2}}\\ \end{array} $

marty cohen
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Write $25$ as $5^2$, $4$ as $2^2$, then only manipulate the exponents as suggested before.

Jan Stout
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