Problem with Indices question

Question

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?

Answer 1

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} $

Answer 2

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