(U.F.Ouro Preto-MG) Resolva a equação logarítmica

por **andersontricordiano** » Ter Set 27, 2011 16:16

Resolva a equação logarítmica:
$({log}_{2}8)({log}_{8}(2-x))+{log}_{2}(1-x)=2+2({log}_{4}3)$

Resposta: S={-2}

Agradeço quem resolver esse calculo!

por **DanielFerreira** » Dom Dez 18, 2011 12:26

andersontricordiano escreveu:Resolva a equação logarítmica:
$({log}_{2}8)({log}_{8}(2-x))+{log}_{2}(1-x)=2+2({log}_{4}3)$

Resposta: S={-2}

Agradeço quem resolver esse calculo!

${log}_{2}2^3 . ({log}_{8}(2 - x)) + {log}_{2}(1 - x) = 2 + {log}_{4}3^2$

$3 . {log}_{8}(2 - x) + {log}_{2}(1 - x) = 2 + {log}_{4}9$

${log}_{8}(2 - x)^3 +{log}_{2}(1 - x) = 2 + {log}_{4}9$

$\frac{{log}_{2}(2 - x)^3}{{log}_{2}8} + {log}_{2}(1 - x) = 2 + \frac{{log}_{2}9}{{log}_{2}4}$

$3\frac{{log}_{2}(2 - x)}{{log}_{2}2^3} + {log}_{2}(1 - x) = 2 + \frac{{log}_{2}3^2}{{log}_{2}2^2}$

$3\frac{{log}_{2}(2 - x)}{3} + {log}_{2}(1 - x) = 2 + 2\frac{{log}_{2}3}{2}$

${log}_{2}(2 - x) + {log}_{2}(1 - x) = 2 + {log}_{2}3$

${log}_{2}[(2 - x)(1 - x)] = 2 + {log}_{2}3$

${log}_{2}(2 - 2x - x - x^2) - {log}_{2}3 = 2$

${log}_{2}\frac{x^2 - 3x + 2}{3} = 2$

$\frac{x^2 - 3x + 2}{3} = 2^2$

x^2 - 3x + 2 = 12