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Math Help - Unknown in Indices

  1. #1
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    Unknown in Indices

    I am having so much trouble with these kinds of equations. Can someone please help me

    Solve for 'x':

    2^2x-1 - 6.2^x-1 + 4 = 0



    and...

    2^2x - 5.2^x+1 + 16 = 0
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  2. #2
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    Quote Originally Posted by Ogsta101 View Post
    I am having so much trouble with these kinds of equations. Can someone please help me

    Solve for 'x':

    2^2x-1 - 6.2^x-1 + 4 = 0



    and...

    2^2x - 5.2^x+1 + 16 = 0
    please clarify (use parentheses!) i'm not sure what the questions are...

    is it:

    2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0

    and

    2^{2x} - 5 \cdot 2^{x + 1} + 16 = 0

    ?
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  3. #3
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    Quote Originally Posted by Ogsta101 View Post
    I am having so much trouble with these kinds of equations. Can someone please help me

    Solve for 'x':

    2^2x-1 - 6.2^x-1 + 4 = 0



    and...

    2^2x - 5.2^x+1 + 16 = 0
    For the first, do you mean 2^{2x - 1} - 6 \times 2^x + 4 = 0 \, ? Otherwise you have 2^{2x - 1} appearing twice .......

    For the second:

    2^{2x} - 5 \times 2^{x+1} + 16 = 0

    \Rightarrow (2^{x})^2 - 5 \times 2 \times 2^{x} + 16 = 0

    \Rightarrow (2^{x})^2 - 10 \times 2^{x} + 16 = 0

    Let w = 2^x:

    \Rightarrow w^2 - 10w + 16 = 0

    \Rightarrow (w - 2)(w - 8) = 0

    \Rightarrow w = 2, \, 8

    Therefore 2 = 2^x \, or 8 = 2^x.

    Therefore 2^1 = 2^x \, or 2^3 = 2^x.

    Therefore x = .......
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  4. #4
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    For the first, I mean:

    <br />
2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0<br />


    Sorry about all the confusion...

    And thanx for help on number 2!
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ogsta101 View Post
    For the first, I mean:

    <br />
2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0<br />


    Sorry about all the confusion...

    And thanx for help on number 2!
    well, it's very similar to what mr fantastic did. same concept. try and see if you can do it
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  6. #6
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    Quote Originally Posted by Ogsta101 View Post
    For the first, I mean:

    <br />
2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0<br />


    Sorry about all the confusion...

    And thanx for help on number 2!
    Just to get you started:

    2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0

    Trick - Multiply through by 2 to make life easier:

    \Rightarrow 2^{2x} - 6 \cdot 2^{x} + 8 = 0

    \Rightarrow (2^{x})^2 - 6 \cdot 2^{x} + 8 = 0
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  7. #7
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    Well, I did it without the multiplying through by 2 and got
    x = 2 or 1
    and they appear to work.




    Thanx heaps for helping me out
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  8. #8
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    Quote Originally Posted by Ogsta101 View Post
    Well, I did it without the multiplying through by 2 and got
    x = 2 or 1
    and they appear to work.




    Thanx heaps for helping me out
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