Question Number 170054 by daus last updated on 15/May/22
Answered by greougoury555 last updated on 15/May/22
$$\:\begin{cases}{{a}=\mathrm{9}^{{x}} }\\{{b}=\mathrm{12}^{{x}} }\\{{a}+{b}=\mathrm{16}^{{x}} }\end{cases}\Rightarrow\mathrm{9}^{{x}} +\mathrm{12}^{{x}} \:=\:\mathrm{16}^{{x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{9}}{\mathrm{12}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{16}}{\mathrm{12}}\right)^{{x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} +\mathrm{1}\:=\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} ;\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\left(\frac{{b}}{{a}}\right)={y} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{y}}\:+\mathrm{1}\:=\:{y}\:;\:{y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\frac{{b}}{{a}} \\ $$