Question

sabendo que os números dois logx e log y estão simultaneamente em PA e PG, calcule x e y.​

Answer 1

{2 , log x , log y}

Vamos começar utilizando a razão da PA:

$log\,y~-~log\,x~=~log\,x~-~2\\\\\\log\,y~-~2log\,x~=~-2\\\\\\log\,y~-~log\,x^2~=~-2\\\\\\log\,\frac{y}{x^2}~=~-2\\\\\\\frac{y}{x^2}~=~10^{-2}\\\\\\y~=~\frac{x^2}{100}\\\\\\\boxed{y~=~\left(\frac{x}{10}\right)^2}$

Vamos agora utilizar a razão da PG:

$\frac{log\,y}{log\,x}~=~\frac{log\,x}{2}\\\\\\2~.~log\,y~=~(log\,x)^2\\\\\\Substituindo~y:\\\\\\2~.~log\,\left(\frac{x}{10}\right)^2~=~(log\,x)^2\\\\\\4.log\frac{x}{10}~=~(log\,x)^2\\\\\\4~.~(log\,x~-~log\,10)~=~(log\,x)^2\\\\\\(log\,x)^2~-~4log\,x~+~4~=~0\\\\\\Substituindo\\w~=~log\,x\\\\\\w^2~-~4w~+~4~=~0\\\\\\Aplicando~Bhaskara\\\\\\\Delta~=~(-4)^2-4.1.4~=~0\\\\\\w'~=~w''~=~\frac{4+\sqrt{0}}{2~.~1}~=~\boxed{2}$

Voltando a substituição feita:

$log\,x~=~w\\\\\\log\,x~=~2\\\\\\x~=~10^2\\\\\\\boxed{x~=~100}\\\\\\\\y~=~\left(\frac{x}{10}\right)^2\\\\\\y~=~\left(\frac{100}{10}\right)^2\\\\\\\boxed{y~=~100}$

A sequencia fica:

{2 , log 100 , log 100}  ou  {2 , 2 , 2}

Como podemos ver, ela é uma PA constante (razão 0) e uma PG constante (razao 1).