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# Indices Help [Factorisation & Simplication] watch

1. Hello, I am new to this website. Can you people provide me the steps to finding the answers to these questions shown below?

1) 4^x - 9

2) 4^x - 25

3) 16 - 9^x

4) 2^n (n+1) + 2^n (n-1)

5) 3^n (n-1/6) - 3^n (n+1/6)

Thank you very much
2. ZXLPK42

Can you please use more brackets as it's not entirely clear what you mean.

Also, the rules of indices will help

3. (Original post by ZXLPK42)
Hello, I am new to this website. Can you people provide me the steps to finding the answers to these questions shown below?

1) 4^x - 9

2) 4^x - 25

3) 16 - 9^x

4) 2^n (n+1) + 2^n (n-1)

5) 3^n (n-1/6) - 3^n (n+1/6)

Thank you very much
You havent actually told us what the question is? There are 5 expressions, what has to be done with each one?
4. (Original post by nerak99)
You havent actually told us what the question is? There are 5 expressions, what has to be done with each one?
First 3 is factorise, last two is simplify.
5. (Original post by Kvothe the arcane)
ZXLPK42

Can you please use more brackets as it's not entirely clear what you mean.

Also, the rules of indices will help

6. (Original post by ZXLPK42)
First 3 is factorise, last two is simplify.
first 3 think about diff of 2 squares, use the fact that to help

last 2 factor out the "power" term first and see what you';re left with, then finish it off
7. (Original post by ZXLPK42)
But and similarly

A few questions seem to want you to use the difference of two squares:

You can factor out something for the last two expressions.
8. (Original post by DylanJ42)
first 3 think about diff of 2 squares, use the fact that to help

last 2 factor out the "power" term first and see what you';re left with, then finish it off
So for example for the first one, do I turn 4^x - 9 into 2^2x - 3^2, and make it equate to (2^x - 3)^2?

And (e.g.) for the fourth one can I turn "2^n (n+1) + 2^n (n-1)" into 2^n (n + 1 + n - 1), and make it so that 2^n (2n)?
9. (Original post by Kvothe the arcane)
But and similarly

A few questions seem to want you to use the difference of two squares:

You can factor out something for the last two expressions.
4x - 9 = 22x - 32 = (2x - 3)2?
10. (Original post by ZXLPK42)
So for example for the first one, do I turn 4^x - 9 into 2^2x - 3^2, and make it equate to (2^x - 3)^2?
No to the bolded. While , .

What you have to notice that they are both squared terms. And there is a difference between them.
11. (Original post by Kvothe the arcane)
No to the bolded. While , .

What you have to notice that they are both squared terms. And there is a difference between them.
So is the final step of the solution 22x - 32?
12. (Original post by ZXLPK42)
So is the final step of the solution 22x - 32?
No because you can simplify it if you know that as I mention above .

(Original post by ZXLPK42)
And (e.g.) for the fourth one can I turn "2^n (n+1) + 2^n (n-1)" into 2^n (n + 1 + n - 1), and make it so that 2^n (2n)?
Yes, but I'm not sure if they'd want you to express that as but what you've said is correct .
13. (Original post by ZXLPK42)
So for example for the first one, do I turn 4^x - 9 into 2^2x - 3^2, and make it equate to (2^x - 3)^2?

And (e.g.) for the fourth one can I turn "2^n (n+1) + 2^n (n-1)" into 2^n (n + 1 + n - 1), and make it so that 2^n (2n)?
into , this is correct, id probably rewrite it further as just to see the squared terms easier

... and make it equate to (2^x - 3)^2? this is not right though

remember that

.
.
.

4th one looks good, although I would factor out another 2
14. (Original post by Kvothe the arcane)
No because you can simplify it if you know that as I mention above .
So does the equation become (2x-3)(2x+3)?
15. (Original post by ZXLPK42)
So does the equation become (2x-3)(2x+3)?
Yes . Although what you have above are expressions and not equations as there is no equals (=) sign. Your meaning is clear, however. I just wanted to point that out.
16. (Original post by DylanJ42)
into , this is correct, id probably rewrite it further as just to see the squared terms easier

... and make it equate to (2^x - 3)^2? this is not right though

remember that

.
.
.

4th one looks good, although I would factor out another 2
How would 2^n(2n) look like if I factored out another 2? I'm not sure if it can be equated as: 2[(2n)^n], but I think it'll be the same as the one above [(2^n(2n)]
17. (Original post by Kvothe the arcane)
Yes . Although what you have above are expressions and not equations as there is no equals (=) sign. Your meaning is clear, however. I just wanted to point that out.
Thank you very much for helping me out.
18. (Original post by ZXLPK42)
How would 2^n(2n) look like if I factored out another 2? I'm not sure if it can be equated as: 2[(2n)^n], but I think it'll be the same as the one above [(2^n(2n)]
remember that means 2 x 2 x 2 x 2 x 2... x 2 'n' times, so multiply that by 2 again and you'll get 2 x 2 x 2 x 2 'n+1' times, so yor answer will be...

Spoiler:
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19. (Original post by DylanJ42)
remember that means 2 x 2 x 2 x 2 x 2... x 2 'n' times, so multiply that by 2 again and you'll get 2 x 2 x 2 x 2 'n+1' times, so yor answer will be...

Spoiler:
Show
I am able to understand the first part, but how does the original value 2n change to n from "2^(n+1) x n"?
20. (Original post by ZXLPK42)
I am able to understand the first part, but how does the original value 2n change to n from "2^(n+1) x n"?
so we have

If you had you could write this as etc etc, ie it doesnt matter what order the letters are in.

Using this, we can rewrite the first expression as

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