Question Number 647 by 123456 last updated on 18/Feb/15
$$\mathrm{log}_{{x}} \left({y}^{\pi} \right)+\mathrm{log}_{{y}} \left({x}^{{e}} \right)={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({x}^{\pi^{−\mathrm{1}} } \right)}−\frac{\mathrm{1}}{\mathrm{log}_{{x}} \left({y}^{{e}^{−\mathrm{1}} } \right)}={b} \\ $$$$\frac{{x}^{{a}+{b}+\mathrm{2}{e}} }{{y}^{{a}−{b}+\mathrm{2}\pi} }=? \\ $$
Commented by 123456 last updated on 17/Feb/15
$$\mathrm{log}_{{x}} \left({y}^{\pi} \right)+\mathrm{log}_{{y}} \left({x}^{{e}} \right)={a} \\ $$$$\pi\mathrm{log}_{{x}} {y}+{e}\mathrm{log}_{{y}} {x}={a} \\ $$$$\mathrm{log}_{{y}} {x}=\frac{\mathrm{log}_{{x}} {x}}{\mathrm{log}_{{x}} {y}}=\frac{\mathrm{1}}{\mathrm{log}_{{x}} {y}} \\ $$$$\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({x}^{\pi^{−\mathrm{1}} } \right)}−\frac{\mathrm{1}}{\mathrm{log}_{{x}} \left({y}^{{e}^{−\mathrm{1}} } \right)}={b} \\ $$$$\frac{\pi}{\mathrm{log}_{{y}} {x}}−\frac{{e}}{\mathrm{log}_{{x}} {y}}={b} \\ $$$$\pi\mathrm{log}_{{x}} {y}−{e}\mathrm{log}_{{y}} {x}={b} \\ $$$${a}+{b}=\left(\pi\mathrm{log}_{{x}} {y}+{e}\mathrm{log}_{{y}} {x}\right)+\left(\pi\mathrm{log}_{{x}} {y}−{e}\mathrm{log}_{{y}} {x}\right)=\mathrm{2}\pi\mathrm{log}_{{x}} {y} \\ $$$${a}−{b}=\left(\pi\mathrm{log}_{{x}} {y}+{e}\mathrm{log}_{{y}} {x}\right)−\left(\pi\mathrm{log}_{{x}} {y}−{e}\mathrm{log}_{{y}} {x}\right)=\mathrm{2}{e}\mathrm{log}_{{y}} {x} \\ $$
Answered by prakash jain last updated on 19/Feb/15
$$\frac{{x}^{{a}+{b}+\mathrm{2}{e}} }{{y}^{{a}−{b}+\mathrm{2}\pi} }=\frac{{x}^{\mathrm{2}\pi\mathrm{log}_{{x}} {y}+\mathrm{2}{e}} }{{y}^{\mathrm{2}{e}\mathrm{log}_{{y}} {x}+\mathrm{2}\pi} }=\frac{{y}^{\mathrm{2}\pi} {x}^{\mathrm{2}{e}} }{{x}^{\mathrm{2}{e}} {y}^{\mathrm{2}\pi} }=\mathrm{1} \\ $$