Quick Indices problem

Question

I'm not entirely sure whether this is a suitable problem to ask. However, in trying to solve an index problem, I have been getting stuck in the same place for a while.

The question is:

$$(9a^3 b^{-4})^{1/2} \times 2(a^{1/2}b^{−2})^{−2}$$

I know that the answer is
$$6a^{1/2} \times b^{2}$$

But I'm unsure exactly how to get there. I know I must start by expanding; however, I then get stuck.

Have you tried to focus on just one factor, for example the powers of $a$? — Paul Sundheim, Sep 23 '17 at 01:40

score 2 · Answer 1 · answered Sep 23 '17 at 01:46

2

Try using these two basic principles:

$(a^x)^y = a^{x \cdot y}$

$a^x \cdot a^y = a^{x+y}$

answered Sep 23 '17 at 01:46

Bram28

100,612
6
70
118

score 1 · Answer 2 · answered Sep 23 '17 at 01:48

1

You have $$(9a^3 b^{-4})^{1/2} \times 2(a^{1/2}b^{−2})^{−2} = 9^{1/2}a^{3/2}b^{-2} \times 2a^{-1}b^{4} = \frac{3a^{3/2}}{b^{2}} \times \frac{2b^4}{a} = 6a^{1/2}b^2$$

answered Sep 23 '17 at 01:48

mate89

538

Oh right thanks, I had expanded my equasion but found myself stuck for the next step, that makes a lot of sense. Thank you – S.Ban Sep 24 '17 at 00:36

score 0 · Answer 3 · answered Aug 05 '19 at 16:35

0

The answer to your question can be viewed via this link: https://i.stack.imgur.com/EKBDO.jpg

answered Aug 05 '19 at 16:35

Nzewi Ernest Kenechukwu

178

Quick Indices problem

3 Answers3