log _4 (5x−6).log


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Question Number 106024 by john santu last updated on 02/Aug/20

$\log_{4} (5 x - 6) . \log_{x} (256) = 8$
Commented by Dwaipayan Shikari last updated on 02/Aug/20

${2, 3}$
	Answered by Dwaipayan Shikari last updated on 02/Aug/20
	$((log(5x−6).log(256))/(log(4).log(x)))=8 ((log(5x−6).8log(2))/(2log(2)log(x)))=8 ((log(5x−6))/(log(x)))=2 5x−6=e^(2log(x)) 5x−6=x^2 x^2 −5x+6=0 { ((x=2)),((x=3)) :}$
	$\frac{l o g (5 x - 6) . l o g (256)}{l o g (4) . l o g (x)} = 8$ $\frac{l o g (5 x - 6) . 8 l o g (2)}{2 l o g (2) l o g (x)} = 8$ $\frac{l o g (5 x - 6)}{l o g (x)} = 2$ $5 x - 6 = e^{2 l o g (x)}$ $5 x - 6 = x^{2}$ $x^{2} - 5 x + 6 = 0$ ${\begin{cases} x = 2 \\ x = 3 \end{cases}$
	Answered by bemath last updated on 02/Aug/20

	$\frac{1}{2} \times 8 \log_{2} (5 x - 6) . \log_{x} (2) = 8$ $\log_{2} (5 x - 6) = 2 . \log_{2} (x)$ $x^{2} - 5 x + 6 = 0$ $x = 2 & x = 3$
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