# Thread: Solving an equation in powers of 10

1. ## Solving an equation in powers of 10

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

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

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

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

\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 $w = 10^{2x^2+4x+2}$:
\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:
\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}

3. Thank you so much! I had tried everything except this.