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C1 Indices Question HELP! Watch

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    Hi,

    Please Could anyone help (with method) how to solve for x (answer is 2)

    2^2x - 6(2^x+1) + 32 = 0
    (2 to the power of 2x minus 6 multiplied by 2 to the power of x+1 plus 32)

    Thanks for the help!
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    (Original post by umbrocurry)
    Hi,

    Please Could anyone help (with method) how to solve for x (answer is 2)

    2^2x - 6(2^x+1) + 32 = 0
    (2 to the power of 2x minus 6 multiplied by 2 to the power of x+1 plus 32)

    Thanks for the help!
    So.... 2^{2x}-6(2^{x+1})+32=0??
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    (Original post by umbrocurry)
    Hi,

    Please Could anyone help (with method) how to solve for x (answer is 2)

    2^2x - 6(2^x+1) + 32 = 0
    (2 to the power of 2x minus 6 multiplied by 2 to the power of x+1 plus 32)

    Thanks for the help!
    You should know that 2^{x+1}=2^x \cdot 2^1

    Your equation becomes: (2^x)^2-6(2)(2^x)+32=0 \Rightarrow (2^x)^2-12(2^x)+32=0 which you can solve for 2^x and then x on its own.

    x=2 is not the only answer.
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    2^{2x}-6(2^{x+1})+32=0

    Remember that  a^{b+c} = a^b \times a^c
    So 2^{x+1} can be written as  2 \times 2^x
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    (Original post by RDKGames)
    You should know that 2^{x+1}=2^x \cdot 2^1

    Your equation becomes: (2^x)^2-6(2)(2^x)+32=0 \Rightarrow (2^x)^2-12(2^x)+32=0 which you can solve for 2^x and then x on its own.

    x=2 is not the only answer.
    Yes but how can I solve as I can't get all the base numbers equal- 32 is 2^5 but then I am left with 12??
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    (Original post by Daffy786)
    Yes but how can I solve as I can't get all the base numbers equal- 32 is 2^5 but then I am left with 12??
    What does that even mean?? Just solve it as you would solve any quadratic - factorise it, or complete the square if needed.
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    (Original post by RDKGames)
    What does that even mean?? Just solve it as you would solve any quadratic - factorise it, or complete the square if needed.
    yes but how exactly?
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    (Original post by umbrocurry)
    yes but how exactly?
    Let y=2^x

    \Rightarrow y^2-12y+32=0

    Then when you have y=something just replace y with 2^x and take logs with base 2 of both sides to get x
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    (Original post by RDKGames)
    Let y=2^x

    \Rightarrow y^2-12y+32=0

    Then when you have y=something just replace y with 2^x and take logs with base 2 of both sides to get x
    Oh Yeah!!! Thanks so much! I forget substitution and I finally got the answer now.

    On a side note: why is (2^x)^2 equal to 2^2x and not 4^2x, since surely you would have to square the integer too?

    Thanks alot
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    (Original post by umbrocurry)
    Oh Yeah!!! Thanks so much! I forget substitution and I finally got the answer now.

    On a side note: why is (2^x)^2 equal to 2^2x and not 4^2x, since surely you would have to square the integer too?

    Thanks alot
    Simple GCSE index rule you're forgetting whereby (a^x)^y=a^{xy}

    It cannot be 4^{2x} since 2^{2x}=2^{x+x}=2^{x} \cdot 2^x=(2^x)^2 which makes no sense as to where the 4 comes from
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    (Original post by RDKGames)
    Index rule you're forgetting whereby (a^x)^y=a^{xy}

    It cannot be 4^{2x} since 2^{2x}=2^{x+x}=2^{x} \cdot 2^x=(2^x)^2 which makes no sense as to where the 4 comes from
    Oh yes of Course! I think it is because i was getting confused between (2^x)^2 and (2x)^2 (which of course equals 4x^2).

    Thanks alot bro
 
 
 
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