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Math Help - I can't find the root(s) of this equation (indices)

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    I can't find the root(s) of this equation (indices)

    Here are my workings solve the equation 2^{2x-1} =16 +2^{x+1} :
    I can't find the root(s) of this equation (indices)-img-1.jpg
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  2. #2
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    Dear PythagorasNeophyte,

    Your quadratic equation is incorrect. It should be, y^2-4y-32=0~where~y=2^x.
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  3. #3
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    Your error was in going from

    2^x2^x2^{-1}=2^4+(2)2^x to 2^x2^x=2^5+(4)2^x

    You inadvertently switched the addition and multiplication symbols on the right-hand side.

    You could also try

    2^{2x-1}-2^{x+1}=2^4

    2^{2x}-2^{x+2}=2^5

    2^x\left[2^x-4\right]=32

    The only two positive numbers that differ by 4 and whose product is 32 are 8 and 4 so by inspection 2^x=8
    Last edited by Archie Meade; December 24th 2010 at 04:22 AM. Reason: emphasis on positive factors of 32
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  4. #4
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    Hello, PythagorasNeophyte!

    Sheesh . . . you're at this level of algebra
    . . and you're still using \times for multiplication?


    2^{2x-1} \:=\:16 +2^{x+1}

    We have: . 2^{2x-1} - 2^{x+1} - 16 \:=\:0

    . . . . . 2^{2x}\cdot2^{-1} - 2^x\cdot 2 - 16 \:=\:0

    . . . . . . . \frac{1}{2}\!\cdot\!2^{2x} - 2\cdot2^x - 16 \:=\:0


    Multiply by 2: . 2^{2x} - 4\cdot2^x - 32 \:=\:0

    Let y = 2^x\!:\;\;y^2 - 4y - 32 \:=\:0\quad\Rightarrow\quad (y + 4)(y-8) \:=\:0 \quad\Rightarrow\quad y \:=\:-4,\:8


    Back-substitute: . \begin{Bmatrix} 2^x \:=\:\text{-}4 & \Rightarrow & \text{no real roots} \\<br />
2^x \:=\:8 & \Rightarrow & \boxed{x \:=\:3} \end{Bmatrix}

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  5. #5
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    If you want a fancy solution
    instead of a quadratic....

    2^{2x-1}=2^4+2^{x+1}

    divide both sides by 2 to obtain 2^x on the right

    2^{2x-2}=2^3+2^x

    2^{2x-2}-2^3=2^x

    Divide both sides by 2^x

    2^{x-2}-2^{3-x}=1

    The only powers of 2 whose index moduli and values differ by 1 are 2^0=1 and 2^1=2

    \Rightarrow\ x=3
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