Simplify and express with positive indices

[jahaddow]

Ok, so I have this Expression, (6^x+2 x 4^2x-4 x 3^5-x x 2^x-6)/(12^4x+3 x 9^2x-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.

 

[lanedance]

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

 

[jahaddow]

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

 

[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.

 

[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.

 

[jahaddow]

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

Thankyou

 

[lanedance]

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

 

[xX-Cyanide-Xx]

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

 

[lanedance]

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

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

 

[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...:)

 

[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

 

[xX-Cyanide-Xx]

i think you've got way more than enough to attempt something...

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

 

[lanedance]

ok so what doesn't make sense?

post #9 is in effect a simple example

 

[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?

 

[xX-Cyanide-Xx]

wait that's not it, never mind.
can u please show me how u did ur working out cus I am completely stuck?

 

[lanedance]

I have in post 9..

part of the confusion may be how its written:

Ok, so I have this Expression, (6^x+2 x 4^2x-4 x 3^5-x x 2^x-6)/(12^4x+3 x 9^2x-3)
QUOTE]

I take it to actually mean the lower crosses are "multlipy", whilst the x's in the exponents are the variable "x", giving below
$$(6^{+2}4^{2x-4} x 3{5-x}x 2{x-6})/(12{4x+3}x 9{2x-3})$$

that's open to interpretation

 

[xX-Cyanide-Xx]

sorry but that's even more confusing. can u show me how u worked it out?
PLEASE!

 

[sponsoredwalk]

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

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.

 

[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?

 

[xX-Cyanide-Xx]

This will sound dumb but i don't even really know wat that means, that's y i wanted the answer, or the working so i can use it withother similar questions.

 

[Tedjn]

I want to clarify, which in particular. Prime factorization, laws of exponents (see https://www.physicsforums.com/showpost.php?p=2759329&postcount=5), or both?

 

[xX-Cyanide-Xx]

Both

 

[sponsoredwalk]

I think you should watch this video

http://khanexercises.appspot.com/video?v=zM_p7tfWvLU

It's number 53 in the list underneath it, if you understand it then watch numbers 54 & 55 after it.

I think it will help you out here a bit.

Come back after you watch these and read people's comments again, I bet everything will be clearer.

Then you should watch the videos on logarithms, ( I think I saw you in another thread with logarithm questions)

 

[xX-Cyanide-Xx]

the video helped a little, but not with this question. it just makes me more fustrated. i can't do it no matter how many time u interpret it.

 

[sponsoredwalk]

If you watched 53,54 & 55 and then look at my first post here that should explain everything.

You've also got about 3 other ways to do this problem just given to you from people on this thread.

I can only guess that you haven't studied this stuff before or haven't looked at a good book explaining it.

There are these notes here = http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx that explicitly taught me how to do these problems ages ago, just read the 'Preliminaries' chapter, but really all you need to do is study this page here http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx that is from that link, I've just checked & the method to do this problem is taught explicitly.

If you give studying this page a shot & still can't do it then I'll show you step by step but you have to give this a shot first to learn for yourself.

I thought the video would just remind you of something you'd learned but you must not have ever seen this before, (or just understood it).

The above chapter I gave you should be 100% all you need to master this problem.

 

[xX-Cyanide-Xx]

i give up, i can't do it.

 

[sponsoredwalk]

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

 

[jahaddow]

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

please Help, I'm really stuck

 

[jahaddow]

sponsoredwalk, that's awesome explaining, but can you work it out for us in just a few steps?