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Thread: problem solving for algebra

  1. #1
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    problem solving for algebra [solved]

    Good day everyone,

    Does anyone know how to solve this,

    Given $\displaystyle 2^{2x}+64(2^{-x})= 32$. Find the value of x.

    Thanks in advance!
    Last edited by imforumer; Jun 22nd 2008 at 05:10 AM. Reason: -
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  2. #2
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    Quote Originally Posted by imforumer View Post
    Given $\displaystyle 2^{2x}+64(2^{-x})= 32$. Find the value of x.
    Welcome to the forum Imforumer

    This first question multiply through the equation by $\displaystyle 2^x$

    $\displaystyle 2^{3x}- 32 \cdot 2^{x}+64=0 $

    let $\displaystyle 2^x = u $

    $\displaystyle u^3 - 32 u+64=0 $

    Notice that $\displaystyle u = 4$ is a solution to this equation.

    Can you finish ?

    Bobak
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  3. #3
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    I also can't solve it, I stuck at where you stuck.

    Btw, the answer given is $\displaystyle x=1.31$(3 significant figure), $\displaystyle x=2$, no solution is provided =.="

    thanks bobak for your time.
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  4. #4
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    Quote Originally Posted by imforumer View Post
    I also can't solve it, I stuck at where you stuck.

    Btw, the answer given is $\displaystyle x=1.31$(3 significant figure), $\displaystyle x=2$, no solution is provided =.="

    thanks bobak for your time.
    If 4 is a root, then $\displaystyle u-4$ is a factor of $\displaystyle u^3 - 32u + 64$.So do polynomial devision and find the quotient.

    $\displaystyle u^3 - 32u + 64 = (u-4)(u^2 + 4u - 16) = 0$

    Surely you should know how to solve the quadratic $\displaystyle u^2 + 4u - 16 = 0$
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