Question

Log2 x + logx 2=5/2. Calcular o valor de x

Answer 1

$log_{2} ^{x} + log_{x} ^{2} = \frac{5}{2} \\ \frac{log(x)}{log(2)} + \frac{log(2)}{log(x)} = \frac{5}{2} \\ \frac{ (log(x))^{2} + (log(2))^{2} }{log(2).log(x)} =\frac{5}{2} \\ \frac{ (log(x))^{2} + (log(2))^{2} }{log(x)} =\frac{5}{2}.log(2)$

Fazendo a mudança:

$log(x) = y$

$\frac{ y^{2} + (log(2))^{2} }{y} =\frac{5}{2}.log(2) \\ \\ 2.(y^{2} + (log(2))^{2}) = 5log(2).y \\ \\ 2y^{2} + 2(log(2))^{2} = 5log(2).y \\ \\ 2y^{2} - 5log(2).y + 2(log(2))^{2} = 0$

Temos uma equação do segundo grau em y:

Δ =(- 5log(2))² - 4 . 2 . 2(log(2))² 
Δ =25.(log(2))² - 16(log(2))²
Δ =9.(log(2))²

$y = \frac{-(- 5log(2))+- \sqrt{9.(log(2))^{2}} }{2.2} \\ \\ y = \frac{5log(2)+- 3log(2) }{4} \\ \\ y_{1} = \frac{5log(2)+ 3log(2) }{4} =\frac{8log(2) }{4} = 2log(2)=log(2^{2}) = log(4) \\ \\ y_{2} = \frac{5log(2)- 3log(2) }{4} =\frac{2log(2) }{4} = \frac{log(2)}{2} = \frac{1}{2} . log(2) = log( 2^{ \frac{1}{2} } ) = log( \sqrt{2} ) \\ \\ Como \\ log(x) = y \\ \\ log(x)=y_{1} \\log(x) = log(4) \\ \boxed{\boxed{x = 4}}$

$log(x)=y_{2} \\log(x) = log( \sqrt{2} ) \\ \boxed{\boxed{x = \sqrt{2} }}$

Logo, 

x = 4 e x = √2