# Simplify expression without using zero or negative indices

• Dec 14th 2008, 06:02 AM
spishak
Simplify expression without using zero or negative indices
Hi! I'm new to the forum, I'm learning maths from some textbooks over the aussie summer holidays and have no teacher or tutor, so I'm in here bothering you guys :)

I have to simplify some expressions without using 0 or a negative exponent, and got to this one:

(2a^-1/3b^-2)^-2

When I worked through it I got 4a^2/9b^4 as the answer, but the textbook says the answer is 9a^2/4b^4. My working went as follows.

(2a^-1/3b^-2)^-2

=

(3b^2/2a)^2

=

(3b^2)^-2/(2a)^-2

=

9b^2*2/4a^-2

=

9b^-4/4a^-2

=

4a^2/9b^-4

I've worked through it a few times and can't figure it out. Can someone tell me where I went wrong? I want to sort it out before I go further.

p.s. I'd use the [tex] codes but they're not showing negative exponents properly when I preview. sorry
• Dec 14th 2008, 06:18 AM
Moo
Hello,
There are many typos or mistakes. I'll correct them in red.
And that will be assuming that what you have to calculate is :
$\left( \frac{2a^{-1}}{3b^{-2}} \right)^{-2}$
(see the code by clicking on it)
Quote:

Originally Posted by spishak
Hi! I'm new to the forum, I'm learning maths from some textbooks over the aussie summer holidays and have no teacher or tutor, so I'm in here bothering you guys :)

I have to simplify some expressions without using 0 or a negative exponent, and got to this one:

(2a^-1/3b^-2)^-2

When I worked through it I got 4a^2/9b^4 as the answer, but the textbook says the answer is 9a^2/4b^4. My working went as follows.

(2a^-1/3b^-2)^-2

=

(2b^2/3a)^-2 << since it's 3b^(-2) and not (3b)^(-2), there is no negative exponent over 2 and 3.

=

(2b^2)^-2/(3a)^-2

=

9b^2*(-2)/4a^-2

=

9b^-4/4a^-2

=

9a^2/4b^+4

I've worked through it a few times and can't figure it out. Can someone tell me where I went wrong? I want to sort it out before I go further.

p.s. I'd use the [tex] codes but they're not showing negative exponents properly when I preview. sorry

Okay, I'll try to write down something that is as near as possible from what you have written.

\begin{aligned}
\left( \frac{2 {\color{blue}a^{-1}}}{3 {\color{blue}b^{-2}}} \right)^{-2}
&= \left( \frac{2 b^2}{3a} \right)^{-2} \\
&= \left( \frac{3a}{2 b^2} \right)^{2} \\
&= \frac{3^2 a^2}{2^2 b^{2*2}} \\
&= \frac{9a^2}{4b^4} \end{aligned}

Does it look clear ? Do tell me if there's something you don't understand !
• Dec 14th 2008, 06:44 AM
spishak

Yes, it's clear! Sorry about the poor syntax, I'm very new to serious maths.

Your interpretation was correct, and I still have to use my way of writing, but if I need to make any more posts after this I will learn and use the markup language the forum uses for my expressions.

As for your correction, i think i got it.

the nature of the term "2a^-1" is such that it could also be written correctly as "2^1*a^-1" and i was interpreting it as "(2a)^-1". i think i get it. thanks a bunch!
• Dec 14th 2008, 06:54 AM
Moo
Quote:

Originally Posted by spishak
Your interpretation was correct, and I still have to use my way of writing, but if I need to make any more posts after this I will learn and use the markup language the forum uses for my expressions.

It comes with the habit ;)

Quote:

the nature of the term "2a^-1" is such that it could also be written correctly as "2^1*a^-1" and i was interpreting it as "(2a)^-1". i think i get it. thanks a bunch!
Exactly ! (Clapping) (Clapping)

Good luck with your studies !