Indices Help [Factorisation & Simplication]

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?

@ZXLPK42

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

Also, the rules of indices will help

First 3 is factorise, last two is simplify.

first 3 think about diff of 2 squares, use the fact that $\displaystyle (a^2)^x = (a^x)^2$ to help

last 2 factor out the "power" term first and see what you';re left with, then finish it off

But $4^x=(2^2)^x$ and similarly $9^x=(3^2)^x$

A few questions seem to want you to use the difference of two squares: $a^2-b^2=(a-b)(a+b)$

You can factor out something for the last two expressions.

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 $4^x-9=2^{2x}-3^2$, $2^{2x}-3^2 \neq (2^x - 3)^2$.

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 2^2x - 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)?

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)?

No because you can simplify it if you know that $a^2-b^2=(a+b)(a-b)$ as I mention above .

So does the equation become (2^x-3)(2^x+3)?

$\displaystyle 4^x - 9$ into $\displaystyle 2^{2x} - 3^2$, this is correct, id probably rewrite it further as $\displaystyle (2^x)^2 - 3^2$ just to see the squared terms easier

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

remember that $\displaystyle a^2 - b^2 = (a+b)(a-b)$

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4th one looks good, although I would factor out another 2

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.

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 $\displaystyle 2^n$ 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...

(this is the answer, so think about it yourself first)

I am able to understand the first part, but how does the original value 2n change to n from "2^(n+1) x n"?