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Indices Help [Factorisation & Simplication] Watch

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    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
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    ZXLPK42

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

    Also, the rules of indices will help

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    (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?
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    (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.
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    (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

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    (Original post by ZXLPK42)
    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
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    (Original post by ZXLPK42)
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    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.
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    (Original post by DylanJ42)
    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
    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)?
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    (Original post by Kvothe the arcane)
    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.
    4x - 9 = 22x - 32 = (2x - 3)2?
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    (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 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.
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    (Original post by Kvothe the arcane)
    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 22x - 32?
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    (Original post by ZXLPK42)
    So is the final step of the solution 22x - 32?
    No because you can simplify it if you know that a^2-b^2=(a+b)(a-b) 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 2^{n+1}n but what you've said is correct .
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    (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)?
     \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)

    .
    .
    .

    4th one looks good, although I would factor out another 2
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    (Original post by Kvothe the arcane)
    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 (2x-3)(2x+3)?
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    (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.
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    (Original post by DylanJ42)
     \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
    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)]
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    (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.
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    (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  \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)
    Spoiler:
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     \displaystyle 2^{n+1} \cdot n
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    (Original post by DylanJ42)
    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)
    Spoiler:
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     \displaystyle 2^{n+1} \cdot n
    I am able to understand the first part, but how does the original value 2n change to n from "2^(n+1) x n"?
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    (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  \displaystyle 2^n (2n)

    If you had  \displaystyle a =bcd you could write this as  \displaystyle a =cbd = dbc = bdc etc etc, ie it doesnt matter what order the letters are in.

    Using this, we can rewrite the first expression as  \displaystyle (2)2^n(n)

     \displaystyle 2 \times 2^n = 2^{n+1}

     \displaystyle \therefore  2 \times 2^n \times n = 2^{n+1} \times n
 
 
 
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