# How do you solve 3^(2x+1) - 26(3^x) - 9 = 0?

Apr 7, 2016

$x = 2$

#### Explanation:

Replace ${3}^{x}$ by $y$ and you will obtain:

$3 {y}^{2} - 26 y - 9$ which can be solved using the quadratic formula:

$y = \frac{26 \pm \sqrt{{26}^{2} - 4 \cdot 3 \cdot \left(- 9\right)}}{2 \cdot 3}$

$y = \frac{26 \pm \sqrt{676 + 108}}{6}$

$y = \frac{26 \pm \sqrt{784}}{6}$

$y = \frac{26 \pm 28}{6}$

$y = \frac{54}{6} = 9 \mathmr{and} y = - \frac{2}{6} = - \frac{1}{3}$

know we solve:

${3}^{x} = 9$ or ${3}^{x} = - \frac{1}{3}$

${3}^{x} = {3}^{2}$ or $n o s o l u t i o n$

$x = 2$