# How do you solve 6^x + 4^x = 9^x?

Apr 17, 2016

$x = \frac{\ln \left(\frac{1 + \sqrt{5}}{2}\right)}{\ln \left(\frac{3}{2}\right)}$

#### Explanation:

Divide by ${4}^{x}$ to form a quadratic in ${\left(\frac{3}{2}\right)}^{x}$.
Use ${6}^{x} / {4}^{x} = {\left(\frac{6}{4}\right)}^{x} = {\left(\frac{3}{2}\right)}^{x} \mathmr{and} {\left(\frac{9}{4}\right)}^{x} = {\left({\left(\frac{3}{2}\right)}^{2}\right)}^{x} = {\left({\left(\frac{3}{2}\right)}^{x}\right)}^{2}$.
${\left({\left(\frac{3}{2}\right)}^{x}\right)}^{2} - {\left(\frac{3}{2}\right)}^{x} - 1 = 0$

So,${\left(\frac{3}{2}\right)}^{x} = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot \left(- 1\right)}}{2} = \frac{1 \pm \sqrt{5}}{2}$

For the positive solution:

${\left(\frac{3}{2}\right)}^{x} = \frac{1 + \sqrt{5}}{2}$

Applying logarythms:

$x \ln \left(\frac{3}{2}\right) = \ln \left(\frac{1 + \sqrt{5}}{2}\right)$

$x = \frac{\ln \left(\frac{1 + \sqrt{5}}{2}\right)}{\ln \left(\frac{3}{2}\right)} = 1.18681439 \ldots .$