Question

Solve $4^x+6^x=9^x$

Answer 1

Dividing by $4^x$ to form a quadratic equation in ${\left(\frac{3}{2}\right)}^x$

$\begin{array}{c}\frac{6^x}{4^x}={\left(\frac{6}{4}\right)}^x={\left(\frac{3}{2}\right)}^{x}and{\left(\frac{9}{4}\right)}^x={\left({\left(\frac{3}{2}\right)}^2\right)}^x={\left({\left(\frac{3}{2}\right)}^x\right)}^2\\ \Rightarrow {\left({\left(\frac{3}{2}\right)}^x\right)}^2-{\left(\frac{3}{2}\right)}^x-1=0\\ \Rightarrow {\left(\frac{3}{2}\right)}^x=\frac{1\pm \sqrt{1-4\cdot 1\cdot \left(-1\right)}}{2}=\frac{1\pm \sqrt{5}}{2}\end{array}$

For +ve solution,

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

Apply log

$\begin{array}{c}x\ln \left(\frac{3}{2}\right)=\ln \left(\frac{1+\sqrt{5}}{2}\right)\\ \Rightarrow x=\frac{\ln \left(\frac{1+\sqrt{5}}{2}\right)}{\ln \left(\frac{3}{2}\right)}=\mathrm{1.18681439....}\end{array}$