solve((8^x −2^x )/(6^x −3^(x ) ))=2. find x


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Question Number 171730 by Mikenice last updated on 20/Jun/22

$s o l v e \frac{8^{x} - 2^{x}}{6^{x} - 3^{x}} = 2 . f i n d x$
	Answered by puissant last updated on 20/Jun/22
	$((8^x −2^x )/(6^x −3^x ))=2 ⇒ ((2^(3x) −2^x )/(2^x .3^x −3^x )) = 2 ⇒ ((2^x (2^x −1)(2^x +1))/(3^x (2^x −1))) = 2 2^x −1 ≠ 0 , x≠ 0 ⇒ ((2^x (2^x +1))/3^x )=2 ⇒ 2^(x−1) (2^x +1)=3^x • for x≥2 , no solution. • for x=1 , 2^(1−1) (2^1 +1)=3=3^1 . • for x≤−1 , no solution. S={ 1 }. ..........Le puissant..........$
	$\frac{8^{x} - 2^{x}}{6^{x} - 3^{x}} = 2 \Rightarrow \frac{2^{3 x} - 2^{x}}{2^{x} . 3^{x} - 3^{x}} = 2$ $\Rightarrow \frac{2^{x} (2^{x} - 1) (2^{x} + 1)}{3^{x} (2^{x} - 1)} = 2$ $2^{x} - 1 \neq 0, x \neq 0$ $\Rightarrow \frac{2^{x} (2^{x} + 1)}{3^{x}} = 2$ $\Rightarrow 2^{x - 1} (2^{x} + 1) = 3^{x}$ $∙ f o r x ⩾ 2, n o s o l u t i o n .$ $∙ f o r x = 1, 2^{1 - 1} (2^{1} + 1) = 3 = 3^{1} .$ $∙ f o r x ⩽ - 1, n o s o l u t i o n .$ $S = {1} .$ $. . . . . . . . . . L e p u i s s a n t . . . . . . . . . .$
	Commented by Mikenice last updated on 20/Jun/22

	$t h a n k s s i r$
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