# Homework Help: Simplify and express with positive indices

1. Jun 13, 2010

Ok, so I have this Expression, (6x+2 x 42x-4 x 35-x x 2x-6)/(124x+3 x 92x-3)

But it needs to be simplifyed, and expressed with positive indices. Now as far as my knowledge takes me, this can't be simplifyed, but then I am probably wrong. If anybody could help to solve, that would be great.

2. Jun 13, 2010

### lanedance

maybe you could start by trying breaking things down into prime components eg.
$$6^{x+2} = 3^{x+2}2^{x+2}$$

3. Jun 13, 2010

I'm sorry but that example made no sense to me at all? I need more explanation.

4. Jun 13, 2010

### lanedance

well say you have, separate powers of 2 multiplied toegther, you can simplify by
$$2^{a}2^{b} = 2^{a+b}$$

so break down each number into its prime components, then multiply similar numbers together
6 = 3.2
4 = 2.2 etc.

5. Jun 13, 2010

### HallsofIvy

The rules of exponents,

$a^xa^y= a^{x+y}$ (which lanedance gave) and $(a^x)^y= a^{xy}$, are usually learned long before taking Calculus!

You will also need to know that $4= 2^2$, $6= 3(2)$, $9= 3^2$, and $12= 2^2 (3)$. Break everything down into powers of 2 and 3 and use the laws of exponents.

6. Jun 13, 2010

Could somebody please do this as an example, this is just the first one in the list I have to do.

Thankyou

7. Jun 13, 2010

### lanedance

you've got all the info, why not try it out?

8. Jun 13, 2010

### xX-Cyanide-Xx

All this made no sence to me either. Can someone show an example?

9. Jun 13, 2010

### lanedance

do for the first terms
$$6^{x+2}.4^{2x-4} = (3.2)^{x+2}.(2.2)^{2x-4} = (3^{x+2}.2^{x+2})(2^{2x-4}. 2^{2x-4}) = (3^{x+2})(2^{x+2}.2^{2x-4}. 2^{2x-4}) = (3^{x+2})(2^{(x+2) + (2x-4) + (2x-4)}) = (3^{x+2})(2^{5x+6})$$

then for terms that are divided, the power will be negative

10. Jun 13, 2010

### xX-Cyanide-Xx

sorry to be a bother but, that really just confused me even more.
If u can could you please show the whole example, and explain everything?
That would help heeps.
THANKS...:)

11. Jun 13, 2010

### lanedance

i think you've got way more than enough to attempt something... even though its not your post, have a try & i'll guide you the through

basically all you need to do is the same for the last 4 terms
$3^{5-x}$ - this is in lowest form
$2^{x-6}$ - this is in lowest form
$\frac{1}{12^{4x+3}} = 12^{-4x-3} = (2.3.3)^{-4x-3}$

and then i've leave the last for you... then multiply all the terms together & collect exponents

12. Jun 13, 2010

### xX-Cyanide-Xx

There may be heeps on info, but it makes no sence to me, thats y i wanted an example so i could attempt questions just like that.

13. Jun 14, 2010

### lanedance

ok so what doesn't make sense?

post #9 is in effect a simple example

14. Jun 14, 2010

### xX-Cyanide-Xx

the whole thing makes sence, can u tell me what answer u got, cus i got 1.5x^3.
Is this correct???

15. Jun 14, 2010

### xX-Cyanide-Xx

wait thats not it, never mind.
can u please show me how u did ur working out cus im completly stuck?

16. Jun 14, 2010

### lanedance

I have in post 9..

part of the confusion may be how its written:

17. Jun 14, 2010

### xX-Cyanide-Xx

sorry but thats even more confusing. can u show me how u worked it out?

18. Jun 14, 2010

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

Is that the equation??? (the dot means multiplication).

Start off by recognizing that you can do things like;

$$6^{(x + 2)} = 6^x \cdot 6^2$$

Play with this one first, compare the values of the Left Hand Side & The Right Hand side by giving different x values, you'll see something surprising, then generalise this idea.

19. Jun 14, 2010

### Tedjn

We can continue to try to make sense of what doesn't make sense to you. Which concept is confusing: prime factorization or the laws of exponents alluded to by HallsofIvy and lanedance?

20. Jun 14, 2010

### xX-Cyanide-Xx

This will sound dumb but i dont even really know wat that means, thats y i wanted the answer, or the working so i can use it withother similar questions.