# Thread: Solving an equation in powers of 10

1. ## Solving an equation in powers of 10

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

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

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

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

\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}
\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}