Question Number 647 by 123456 last updated on 18/Feb/15
![log_x (y^π )+log_y (x^e )=a (1/(log_y (x^π^(−1) )))−(1/(log_x (y^e^(−1) )))=b (x^(a+b+2e) /y^(a−b+2π) )=?](https://www.tinkutara.com/question/Q647.png)
$$\mathrm{log}_{{x}} \left({y}^{\pi} \right)+\mathrm{log}_{{y}} \left({x}^{{e}} \right)={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({x}^{\pi^{−\mathrm{1}} } \right)}−\frac{\mathrm{1}}{\mathrm{log}_{{x}} \left({y}^{{e}^{−\mathrm{1}} } \right)}={b} \\ $$$$\frac{{x}^{{a}+{b}+\mathrm{2}{e}} }{{y}^{{a}−{b}+\mathrm{2}\pi} }=? \\ $$
Commented by 123456 last updated on 17/Feb/15
![log_x (y^π )+log_y (x^e )=a πlog_x y+elog_y x=a log_y x=((log_x x)/(log_x y))=(1/(log_x y)) (1/(log_y (x^π^(−1) )))−(1/(log_x (y^e^(−1) )))=b (π/(log_y x))−(e/(log_x y))=b πlog_x y−elog_y x=b a+b=(πlog_x y+elog_y x)+(πlog_x y−elog_y x)=2πlog_x y a−b=(πlog_x y+elog_y x)−(πlog_x y−elog_y x)=2elog_y x](https://www.tinkutara.com/question/Q648.png)
$$\mathrm{log}_{{x}} \left({y}^{\pi} \right)+\mathrm{log}_{{y}} \left({x}^{{e}} \right)={a} \\ $$$$\pi\mathrm{log}_{{x}} {y}+{e}\mathrm{log}_{{y}} {x}={a} \\ $$$$\mathrm{log}_{{y}} {x}=\frac{\mathrm{log}_{{x}} {x}}{\mathrm{log}_{{x}} {y}}=\frac{\mathrm{1}}{\mathrm{log}_{{x}} {y}} \\ $$$$\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({x}^{\pi^{−\mathrm{1}} } \right)}−\frac{\mathrm{1}}{\mathrm{log}_{{x}} \left({y}^{{e}^{−\mathrm{1}} } \right)}={b} \\ $$$$\frac{\pi}{\mathrm{log}_{{y}} {x}}−\frac{{e}}{\mathrm{log}_{{x}} {y}}={b} \\ $$$$\pi\mathrm{log}_{{x}} {y}−{e}\mathrm{log}_{{y}} {x}={b} \\ $$$${a}+{b}=\left(\pi\mathrm{log}_{{x}} {y}+{e}\mathrm{log}_{{y}} {x}\right)+\left(\pi\mathrm{log}_{{x}} {y}−{e}\mathrm{log}_{{y}} {x}\right)=\mathrm{2}\pi\mathrm{log}_{{x}} {y} \\ $$$${a}−{b}=\left(\pi\mathrm{log}_{{x}} {y}+{e}\mathrm{log}_{{y}} {x}\right)−\left(\pi\mathrm{log}_{{x}} {y}−{e}\mathrm{log}_{{y}} {x}\right)=\mathrm{2}{e}\mathrm{log}_{{y}} {x} \\ $$
Answered by prakash jain last updated on 19/Feb/15
![(x^(a+b+2e) /y^(a−b+2π) )=(x^(2πlog_x y+2e) /y^(2elog_y x+2π) )=((y^(2π) x^(2e) )/(x^(2e) y^(2π) ))=1](https://www.tinkutara.com/question/Q649.png)
$$\frac{{x}^{{a}+{b}+\mathrm{2}{e}} }{{y}^{{a}−{b}+\mathrm{2}\pi} }=\frac{{x}^{\mathrm{2}\pi\mathrm{log}_{{x}} {y}+\mathrm{2}{e}} }{{y}^{\mathrm{2}{e}\mathrm{log}_{{y}} {x}+\mathrm{2}\pi} }=\frac{{y}^{\mathrm{2}\pi} {x}^{\mathrm{2}{e}} }{{x}^{\mathrm{2}{e}} {y}^{\mathrm{2}\pi} }=\mathrm{1} \\ $$