Question Number 66396 by hmamarques1994@gmai.com last updated on 14/Aug/19
![Seja 53^(log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ]) = 1. Calcule (x_1 /x_2 )+0,9.](Q66396.png)
$$\: \\ $$$$\:\boldsymbol{\mathrm{Seja}}\:\:\mathrm{53}^{\boldsymbol{\mathrm{log}}_{\frac{\mathrm{1}}{\sqrt{\boldsymbol{{e}}^{\boldsymbol{\pi}} }}} \left[\sqrt[{\mathrm{9999999}}]{\left(\boldsymbol{{x}}+\mathrm{11}\right)!}\right]} \:=\:\mathrm{1}. \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Calcule}}\:\:\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} }{\boldsymbol{\mathrm{x}}_{\mathrm{2}} }+\mathrm{0},\mathrm{9}. \\ $$
Answered by mr W last updated on 14/Aug/19
![53^(log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ]) = 1 ⇒log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ]=0 ⇒(((x+11)!))^(1/(9999999)) =1 ⇒(x+11)!=1 ⇒x+11=0 or 1 ⇒x=−11 or −10 ⇒x_1 =−11, x_2 =−10 or x_1 =−10, x_2 =−11 ⇒(x_1 /x_2 )+0.9=((11)/(11))+(9/(10))=2 or ⇒(x_1 /x_2 )+0.9=((10)/(11))+(9/(10))=((199)/(110))](Q66397.png)
$$\mathrm{53}^{\boldsymbol{\mathrm{log}}_{\frac{\mathrm{1}}{\sqrt{\boldsymbol{{e}}^{\boldsymbol{\pi}} }}} \left[\sqrt[{\mathrm{9999999}}]{\left(\boldsymbol{{x}}+\mathrm{11}\right)!}\right]} \:=\:\mathrm{1} \\ $$$$\Rightarrow\boldsymbol{\mathrm{log}}_{\frac{\mathrm{1}}{\sqrt{\boldsymbol{{e}}^{\boldsymbol{\pi}} }}} \left[\sqrt[{\mathrm{9999999}}]{\left(\boldsymbol{{x}}+\mathrm{11}\right)!}\right]=\mathrm{0} \\ $$$$\Rightarrow\sqrt[{\mathrm{9999999}}]{\left(\boldsymbol{{x}}+\mathrm{11}\right)!}=\mathrm{1} \\ $$$$\Rightarrow\left({x}+\mathrm{11}\right)!=\mathrm{1} \\ $$$$\Rightarrow{x}+\mathrm{11}=\mathrm{0}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow{x}=−\mathrm{11}\:{or}\:−\mathrm{10} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =−\mathrm{11},\:{x}_{\mathrm{2}} =−\mathrm{10}\:{or}\:{x}_{\mathrm{1}} =−\mathrm{10},\:{x}_{\mathrm{2}} =−\mathrm{11} \\ $$$$\Rightarrow\frac{{x}_{\mathrm{1}} }{{x}_{\mathrm{2}} }+\mathrm{0}.\mathrm{9}=\frac{\mathrm{11}}{\mathrm{11}}+\frac{\mathrm{9}}{\mathrm{10}}=\mathrm{2} \\ $$$${or} \\ $$$$\Rightarrow\frac{{x}_{\mathrm{1}} }{{x}_{\mathrm{2}} }+\mathrm{0}.\mathrm{9}=\frac{\mathrm{10}}{\mathrm{11}}+\frac{\mathrm{9}}{\mathrm{10}}=\frac{\mathrm{199}}{\mathrm{110}} \\ $$