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Math Help - Solving an equation in powers of 10

  1. #1
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    Solving an equation in powers of 10

    Solve  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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     10^{(x+1)(3x+4)} - 2*10^{(x+1)(x+2)} = 10^{1-x-x^2}

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

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

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

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

    Quadratic formula on w:
    \begin{aligned}<br />
w &= \frac{2 \pm \sqrt{4 + 40}}{2} \\<br />
w &= \frac{2 \pm \sqrt{44}}{2} \\<br />
w &= \frac{2 \pm 2\sqrt{11}}{2} \\<br />
w &= 1 \pm \sqrt{11}<br />
\end{aligned}

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

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