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Math Help - Solve for x...

  1. #1
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    Solve for x...

    Hi,

    Solve for x:

    (a.)\,4^x - 13(2^x) + 40 = 0

    This is supposed to have two solutions (quadratic), but I've only progressed two steps. I believe these two steps to be incorrect. Argh!

    4^x - 26^x + 40 = 0
    -22^x + 40 = 0
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  2. #2
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    They are incorrect.

    13(2^x)\neq 26^x and 4^x - 26^x \neq -22^x

    Recall BODMAS and 'adding like terms' for reasons why.

    Try this

    \displaystyle 4^x - 13(2^x) + 40 = 0

    \displaystyle (2^2)^x - 13(2^x) + 40 = 0

    \displaystyle (2)^{2x} - 13(2^x) + 40 = 0

    \displaystyle (2^x)^{2} - 13(2^x) + 40 = 0

    Now making 2^x = a you get

    \displaystyle a^{2} - 13a + 40 = 0

    Can you continue? What do you get?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Can you continue? What do you get?
    Thanks.

    (a - 5)(a - 8) = 0
    a = 5; a = 8

    2^x\,=\,5; 2^x\,=\,8

    \log(2^x) = \log\,5;\,\,\,\,\,\,\,\,\,\log\,(2^x) = \log\,8

    x = \frac{\log\,5}{\log\,2};\,\,\,\,\,x = \frac{\log\,8}{\log\,2}

    x \approx 2.32;\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3

    13(2^x)\neq 26^x\,\,\,\,\,\,\,\,\,\,\,4^x - 26^x \neq -22^x
    'Like' here means: of the same base?
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  4. #4
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    Quote Originally Posted by Hellbent View Post

    13(2^x)\neq 26^x\,\,\,\,\,\,\,\,\,\,\,4^x - 26^x \neq -22^x
    'Like' here means: of the same base?
    Yep, like this

    4^x - (5)4^x = -(4)4^x
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