# Simplify and express with positive indices

i give up, i cant do it.

The original equation:

$$\frac{6^{(x + 2)} \ \cdot \ 4^{(2x - 4)} \ \cdot \ 3^{(5 - x)} \ \cdot \ 2^{(x - 6)}}{12^{(4x + 3)} \ \cdot \ 9^{(2x - 3)}}$$

Break up the terms by the laws of exponents:

$$\frac{ [6^x \cdot 6^2] \ \cdot \ [4^{2x} \ \cdot \ 4^{(- 4)}] \ \cdot \ [3^5 \ \cdot \ 3^{(- x)}] \ \cdot \ [2^x \ \cdot2^{(- 6)}]}{[12^{4x} \ \cdot \ 12^3] \ \cdot \ [ 9^{(2x)} \ \cdot \ 9^{(- 3)}]}$$

I'm going to put the 4 on top of the fraction that has a minus 4 exponent down to the bottom of the fraction.

this is allowed by the laws of exponents.

$$\frac{ [6^x \cdot 6^2] \ \cdot \ 4^{2x} \ \cdot \ [3^5 \ \cdot \ 3^{(- x)}] \ \cdot \ [2^x \ \cdot2^{(- 6)}]}{[12^{4x} \ \cdot \ 12^3] \ \cdot \ [ 9^{(2x)} \ \cdot \ 9^{(- 3)}] \ \cdot \ 4^{(- 4)}}$$

Do you see where it is? Well, I actually have to change this to a plus four in the exponent, I'll do it now:

$$\frac{ [6^x \cdot 6^2] \ \cdot \ 4^{2x} \ \cdot \ [3^5 \ \cdot \ 3^{(- x)}] \ \cdot \ [2^x \ \cdot2^{(- 6)}]}{[12^{4x} \ \cdot \ 12^3] \ \cdot \ [ 9^{(2x)} \ \cdot \ 9^{(- 3)}] \ \cdot \ 4^{(4)}}$$

See, now it's legal. I just left it to show you, okay, now I'll do the same thing with the 2 hat has a minus 6.

$$\frac{ [6^x \cdot 6^2] \ \cdot \ 4^{2x} \ \cdot \ [3^5 \ \cdot \ 3^{(- x)}] \ \cdot \ 2^x }{[12^{4x} \ \cdot \ 12^3] \ \cdot \ [ 9^{(2x)} \ \cdot \ 9^{(- 3)}] \ \cdot \ 4^ 4 \ \cdot2^ 6}$$

Now with the 3 to the minus x power as well, I'm also going to move the 9 with a minus 3 that is on the bottom up to the top too:

$$\frac{ [6^x \cdot 6^2] \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{[12^{4x} \ \cdot \ 12^3] \ \cdot \ 9^{(2x)} \ \cdot \ 4^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

Okay, now I have no minus signs in the exponent. I'll try and get everything into 2's and 3's so that I can cancel things after I make it look nice:

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 12^3 \ \cdot \ 9^{(2x)} \ \cdot \ 4^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

There, okay I can rewrite the 12^(4x) as [3 x 4]^(4x), if you don't believe me
just practice this yourself, forget the x and just practice with 12^4 = (4^4)x(3^4)
or better yet, get a calculator and see if 12^4 - [(4^4)x(3^4)] = 0, I bet it does ;)

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ (4 \cdot 3)^3 \ \cdot \ 9^{(2x)} \ \cdot \ 4^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

Okay, read this carefully & look out for what I do over the next few times, you'll see what I'm doing;

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 4^3 \cdot 3^3 \ \cdot \ 9^{(2x)} \ \cdot \ 4^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ (2 \cdot 2)^3 \cdot 3^3 \ \cdot \ 9^{(2x)} \ \cdot \ 4^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \cdot 2^3 \cdot 3^3 \ \cdot \ 9^{(2x)} \ \cdot \ (2 \cdot 2)^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \cdot 2^3 \cdot 3^3 \ \cdot \ 9^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \cdot 2^3 \cdot 3^3 \ \cdot \ (3 \cdot 3)^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \cdot 2^3 \cdot 3^3 \ \cdot \ 3^{(2x)} \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot2^ 6 \cdot \ 3^x}$$

$$\frac{ 6^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ (3 \cdot 2)^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ 6^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ (3 \ \cdot \ 2)^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 9^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ (3 \cdot \ 3)^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ 3^5 \ \cdot \ 2^x \cdot \ 3^3 \cdot \ 3^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

Now, the 3^5 can be rewritten as (3^3)x(3^2), if you don't believe me play with it.

$$\frac{ 3^x \cdot \ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ [3^3 \ \cdot \ 3^2] \ \cdot \ 2^x \cdot \ 3^3 \cdot \ 3^3}{12^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ 3^3 \ \cdot \ 3^2 \ \cdot \ 2^x \cdot \ 3^3 \cdot \ 3^3}{(3 \ \cdot \ 4)^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

$$\frac{ 3^x \cdot \ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ 3^3 \ \cdot \ 3^2 \ \cdot \ 2^x \cdot \ 3^3 \cdot \ 3^3}{3^{4x} \ \cdot \ 4^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \cdot \ 3^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6 \cdot \ 3^x}$$

Now I'll cancel some of the things on top and bottom, 3^3 and 3^x;

$$\frac{\ 2^x \cdot \ 3^2 \ \cdot \ 2^2 \ \cdot \ 4^{2x} \ \cdot \ 3^3 \ \cdot \ 3^2 \ \cdot \ 2^x \cdot \ 3^3}{3^{4x} \ \cdot \ 4^{4x} \ \cdot \ 2^3 \ \cdot 2^3 \ \cdot \ 3^{(2x)} \ \cdot 3^{(2x)} \ \cdot \ 2^4 \cdot 2^ 4 \ \cdot \ 2^ 6}$$

and I would keep going except that I have to go, I'm sure you've got the idea anyway. You want it as simplified as possible.

Btw: these get really easy & I went way overboard here, these questions take about 10 seconds to do when you get the hang of them. You want them as factored as possible, there's more that can be done & all of this could of been done quicker but I wanted to show you what you could have done & to give you ideas.

Good luck with it :p

the lower case x's with spaces either side are multilpication signs