# Solve the following ?

## ${4}^{x} + {6}^{x} = {9}^{x}$

Jun 25, 2018

${4}^{x} + {6}^{x} = {9}^{x}$

$\implies {4}^{x} / {6}^{x} + 1 = {9}^{x} / {6}^{x}$

$\implies {\left(\frac{4}{6}\right)}^{x} + 1 = {\left(\frac{9}{6}\right)}^{x}$

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

Let ${\left(\frac{3}{2}\right)}^{x} = y$

So $x \log \left(\frac{3}{2}\right) = \log y$

$\implies x = \frac{\log y}{\log} 1.5$

Again

$y - \frac{1}{y} = 1$

$\implies {y}^{2} - y - 1 = 0$

$\implies y = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \left(- 1\right)}}{2}$

$\implies y = \frac{1 \pm \sqrt{5}}{2}$

So to get the value of $x$ we can use the positive root of $y$ as $x$ is related with $\log y$.

Hence $x = \log \frac{\frac{\sqrt{5} + 1}{2}}{\log} 1.5$