Find all real values $x$ such that $4^x+6^x=9^x$.🔑
Solutions: Dividing $6^x$ on both side of the equation, we have: $$\Big{(}\dfrac{2}{3}\Big{)}^x+1=\Big{(}\dfrac{3}{2}\Big{)}^x$$
Let $y=\Big{(}\dfrac{2}{3}\Big{)}^x$, we have $$y+1=\dfrac{1}{y}$$ i.e. $$y^2+y-1=0$$
Solving the above equation, we have $$y=\dfrac{-1\pm\sqrt{5}}{2}$$ Because $x$ is real, and $y=\Big{(}\dfrac{2}{3}\Big{)}^x>0$, ignoring the negative $y$ value, we have $$\Big{(}\dfrac{2}{3}\Big{)}^x=\dfrac{\sqrt{5}-1}{2}$$ Solving the above equation, we have $$x=\dfrac{\text{ln}\Big{(}\dfrac{\sqrt{5}-1}{2}\Big{)}}{\text{ln}\dfrac{2}{3}}=\boxed{\dfrac{\text{ln}(\sqrt{5}-1)-\text{ln}(2)}{\text{ln}(2)-\text{ln}(3)}}$$