# Math Help - help with algebraic indices ?

1. ## help with algebraic indices ?

Hi,
First time post, so forgive me if my work is a mess I hope you understand it.
Heres the question I have, I think I have it right, would just like someone to check it over.

$\frac{8*2^3*3^2}{4*4^{-2}}$

numerator
${2^3*2^3*3^2}$

${2^6*3^2}$

denominator

${2^2*4^{-2}}$

I bring the demoninator up to the top so the negative and positives change.

${2^{-2}}*{2^2*2^2}$

${2^6*3^2*2^{-2}}*{2^2*2^2}$

${2^8*3^2}$

2. I'm sorry, I have no idea what you're asking here.

3. Sorry I'm trying to simplify this question.
Maybe its in the wrong forum.

$\frac{8*2^3*3^2}{4*4^{-2}}
$

4. Hello !

You want to simplify the following :

$\frac{8 \times 2^3 \times 3^2}{4 \times 4^{-2}}$

Now, the trap to avoid is to not simplify the $8$ and the $4$ together ! They will help, look, it becomes :

$\frac{2^3 \times 2^3 \times 3^2}{4^1 \times 4^{-2}}$

You can combine some indices :

$\frac{2^6 \times 3^2}{4^{-1}}$

Now, note that $2^6 = (2^2)^3 = 4^3$ (laws of indices). Therefore, your expression becomes :

$\frac{4^3 \times 3^2}{4^{-1}}$

Now you can apply the laws of indices to simplify the fours :

$\frac{4^4 \times 3^2}{1}$

That is :

$4^4 \times 3^2$

So :

$\frac{8 \times 2^3 \times 3^2}{4 \times 4^{-2}} = 4^4 \times 3^2 \ \ \ (= 2304)$

The trick here was not to simplify the obvious first, otherwise the possible simplification of the problem would have become conceptually far more difficult to achieve. Use what you are given, the numbers must be there for a reason

But your solution is correct as well, it is a different approach, and you find the same thing ( $2^8 = 4^4$)