Question

utilizando mudança de base e o artificio de chamar
$log_{3}(x = y)$
determine a solução da equação logarítmica
$2 (log_{3}x)^{2} - log_{9}(x) - log_{81}(3) = 0$

Answer 1

Mudando as bases

$2(\log_3x)^2-{\log_3x\over\log_39}-{\log_33\over\log_381}=0\\ \\ 2(\log_3x)^2-{\log_3x\over2}-{1\over4}=0\\ \\ Se~~\log_3x=y\\ \\ 2(y)^2-{y\over2}-{1\over4}=0\\ \\ mmc(2,4)=4\\ \\8y^2-2y-1=0$

a=8

b=-2

c=-1

Δ=b²-4ac

Δ=(-2)²-4(8)(-1)

Δ=4+32

Δ=36

$y={-b\pm\sqrt{\Delta} \over2a}=~~{-(-2)\pm\sqrt{36} \over2(8)}={2\pm6\over16}\\ \\ y'={2+6\over16}={8\over16}={1\over2}\\ \\ y"={2-6\over16}=-{4\over16}=-{1\over4}$

Calculando x

$p/~~y'={1\over2}\\ \\ \log_3x'={1\over2}\\ \\ x'=3^{1\over2}\\ \\ x=\sqrt{3} \\ \\ para~~y"=-{1\over4}\\ \\ \log_3x=-{1\over4}\\ \\ x=3^{-{1\over4}}\\ \\ x=\sqrt[4]{3^{-1}} \\ \\ x=\sqrt[4]{{1\over3}}$