Question

Resolva o sistema:
$\left \{ {{log_{2}x + log_{4}y = 4} \atop {x \times y = 8}} \right.$

Answer 1

$( log_{2}(x) + log_{4}(y) = 4 \\ (x.y = 8 \\ \\ ( log_{2}(x) + log_{2^{2} }(y) = 4 \\ (x.y = 8 \\ \\ ( log_{2}(x) + \frac{1}{2} . log_{2}(y) = 4 \\ (x.y = 8 \\ \\ (2 log_{2}(x) + log_{2}(y) = 8 \\ (x.y = 8 \\ \\ ( log_{2}(x^{2} ) + log_{2}(y) = 8 \\ (x.y = 8 \\ \\ ( log_{2}((x^{2} y) = 8 \\ (x.y = 8 \\ \\ (x^{2} y = 2^{8} \\ (x.y = 8 \\ \\ (x^{2} y = 256 \\ (x.y = 8 \\ \\ (xy) = (32 e \frac{1}{4} )$

$verificando \\ \\ \\ log_{2}(32) + log_{4}( \frac{1}{4} ) = 4 \\ 32 \times \frac{1}{4} = 8 \\ \\ logo \\ \\ 4 = 4 \: e \: 8 = 8$

Answer 2

Propriedades utilizadas:

$log_{a^b}(x) = {log_a(x) \over b} \\\\ log_a(b) + log_a(c) = log_a(bc) \\\\ clog_a(b) = log_a(b^c)$

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$\begin{cases} log_2(x) + log_4(y) = 4 \: (i) \\\\ xy = 8 \: (ii) \end{cases}$

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Simplifique a primeira equação ( i ):

$log_2(x) + log_4(y) = 4 \\\\ log_2(x) = log_{2^2}(y) = 4 \\\\ log_2(x) + { log_2(y) \over 2} = 4 \\\\ 2log_2(x) + log_2(y) = 8 \\\\ log_2(x^2) + log_2(y) = 8 \\\\ log_2(x^2y) = 8 \\\\ x^2y = 2^8 \\\\ x^2y = 256$

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Divida a primeira equação ( i ) pela segunda ( ii ):

${ x^2y \over x y } = { 256 \over 8} \\\\ x = 32$

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Substitua x em ( ii ):

$32y = 8 \\\\ y = { 8 \over 32} \\\\ y = { 1 \over 4}$

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Solução:

$\boxed{\boxed{(x \: , \: y ) = \left( 32 \: , \: {1 \over 4} \right)}}$