Question Number 68712 by Rio Michael last updated on 15/Sep/19
![given that x and y are two numbers other one. given that a>0 and b>0 and a^x = b^y = (ab)^(xy) show that x + y =0](https://www.tinkutara.com/question/Q68712.png)
$${given}\:{that}\:{x}\:{and}\:{y}\:{are}\:{two}\:{numbers}\:{other}\:{one}.\: \\ $$$${given}\:{that}\:\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$${and}\:\:{a}^{{x}} \:=\:{b}^{{y}} \:=\:\left({ab}\right)^{{xy}} \:\:{show}\:{that}\:\:{x}\:+\:{y}\:=\mathrm{0} \\ $$
Commented by Prithwish sen last updated on 15/Sep/19
![Let a^x =b^y =(ab)^(xy) =k ∴ a=k^(1/x) b=k^(1/y) ab=k^(1/(xy)) ∴k^(1/x) .k^(1/y) =k^(1/(xy)) ⇒(1/x)+(1/y) = (1/(xy))⇒x+y=1 please check](https://www.tinkutara.com/question/Q68715.png)
$$\mathrm{Let}\:\mathrm{a}^{\mathrm{x}} =\mathrm{b}^{\mathrm{y}} =\left(\mathrm{ab}\right)^{\mathrm{xy}} =\boldsymbol{\mathrm{k}} \\ $$$$\therefore\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{k}}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \\ $$$$\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{k}}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}} \\ $$$$\boldsymbol{\mathrm{ab}}=\boldsymbol{\mathrm{k}}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{xy}}}} \\ $$$$\therefore\mathrm{k}^{\frac{\mathrm{1}}{\mathrm{x}}} .\mathrm{k}^{\frac{\mathrm{1}}{\mathrm{y}}} =\mathrm{k}^{\frac{\mathrm{1}}{\mathrm{xy}}} \Rightarrow\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{xy}}}\Rightarrow\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$
Commented by MJS last updated on 15/Sep/19
![a^0 +b^0 =(ab)^0 ⇒ x=y=0 is one solution a^x =b^y ⇒ y=((ln a)/(ln b))x a^x =(ab)^(xy) ⇒ y=((ln a)/(ln a +ln b)) ((ln a)/(ln b))x=((ln a)/(ln a +ln b)) ⇒ x=((ln b)/(ln a +ln b)) ⇒ x+y=1](https://www.tinkutara.com/question/Q68741.png)
$${a}^{\mathrm{0}} +{b}^{\mathrm{0}} =\left({ab}\right)^{\mathrm{0}} \:\Rightarrow\:{x}={y}=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{one}\:\mathrm{solution} \\ $$$${a}^{{x}} ={b}^{{y}} \:\Rightarrow\:{y}=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{b}}{x} \\ $$$${a}^{{x}} =\left({ab}\right)^{{xy}} \:\Rightarrow\:{y}=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{a}\:+\mathrm{ln}\:{b}} \\ $$$$\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{b}}{x}=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{a}\:+\mathrm{ln}\:{b}}\:\Rightarrow\:{x}=\frac{\mathrm{ln}\:{b}}{\mathrm{ln}\:{a}\:+\mathrm{ln}\:{b}} \\ $$$$\Rightarrow\:{x}+{y}=\mathrm{1} \\ $$
Commented by Prithwish sen last updated on 15/Sep/19
![Thanks Sir.](https://www.tinkutara.com/question/Q68756.png)
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$