Question Number 222309 by fantastic last updated on 22/Jun/25
$$\begin{array}{|c|}{\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} }\\\hline\end{array} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Jun/25
$$\frac{\mathrm{4}^{{x}} }{\mathrm{6}^{{x}} }+\mathrm{1}=\frac{\mathrm{9}^{{x}} }{\mathrm{6}^{{x}} } \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \\ $$$${let}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} ={y} \\ $$$${y}+\mathrm{1}=\frac{\mathrm{1}}{{y}} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)_{>\mathrm{0}} ^{{x}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}\mathrm{log}\:\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{log}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${x}=\frac{\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{5}}\right)\:−\:\mathrm{log}\:\mathrm{2}\:}{\mathrm{log}\:\mathrm{2}\:−\mathrm{log}\:\mathrm{3}} \\ $$
Commented by fantastic last updated on 22/Jun/25
$$\mathbb{N}{i}\mathbb{C}\epsilon! \\ $$
Answered by Rasheed.Sindhi last updated on 22/Jun/25
$$\frac{\mathrm{4}^{{x}} }{\mathrm{9}^{{x}} }+\frac{\mathrm{6}^{{x}} }{\mathrm{9}^{{x}} }=\mathrm{1} \\ $$$$\:\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}{x}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{3}} \\ $$$$\:\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)_{>\mathrm{0}} ^{{x}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\:\:\:{Further}\:{as}\:{above} \\ $$