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Thread: Solving an equation in powers of 10

  1. #1
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    Solving an equation in powers of 10

    Solve $\displaystyle 10^{(x+1)(3x+4)} - 2*10^{(x+1)(x+2)} = 10^{1-x-x^2} $
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  2. #2
    Senior Member eumyang's Avatar
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    $\displaystyle 10^{(x+1)(3x+4)} - 2*10^{(x+1)(x+2)} = 10^{1-x-x^2} $

    $\displaystyle 10^{3x^2+7x+4} - 2*10^{x^2+3x+2} = 10^{-x^2-x+1} $

    Multiply both sides by $\displaystyle 10^{x^2 + x}$:
    $\displaystyle 10^{4x^2+8x+4} - 2*10^{2x^2+4x+2} = 10$

    $\displaystyle 10^{2(2x^2+4x+2)} - 2*10^{2x^2+4x+2} = 10$

    Let $\displaystyle w = 10^{2x^2+4x+2}$:
    $\displaystyle w^2 - 2w = 10$
    $\displaystyle w^2 - 2w - 10 = 0$

    Quadratic formula on w:
    $\displaystyle \begin{aligned}
    w &= \frac{2 \pm \sqrt{4 + 40}}{2} \\
    w &= \frac{2 \pm \sqrt{44}}{2} \\
    w &= \frac{2 \pm 2\sqrt{11}}{2} \\
    w &= 1 \pm \sqrt{11}
    \end{aligned}$

    Substitute back $\displaystyle w = 10^{2x^2+4x+2}$:
    $\displaystyle \begin{aligned}
    10^{2x^2+4x+2} &= 1 \pm \sqrt{11} \\
    log (10^{2x^2+4x+2}) &= log (1 \pm \sqrt{11}) \\
    2x^2+4x+2 &= log (1 \pm \sqrt{11})
    \end{aligned}$

    We can ignore 1 - sqrt(11) because it's negative. Moving on:
    $\displaystyle \begin{aligned}
    2x^2+4x+2 &= log (1 + \sqrt{11}) \\
    x^2 + 2x +1 &= \frac{log (1 + \sqrt{11})}{2} \\
    (x + 1)^2 &= \frac{log (1 + \sqrt{11})}{2} \\
    x + 1 &= \pm \sqrt{\frac{log (1 + \sqrt{11})}{2}} \\
    x &= -1 \pm \sqrt{\frac{log (1 + \sqrt{11})}{2}} \\
    \end{aligned}$
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  3. #3
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    Thank you so much! I had tried everything except this.
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