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Question Number 200696 by Bayat last updated on 22/Nov/23

	Answered by aleks041103 last updated on 22/Nov/23

	${(2^{x} - 3^{x})}^{2} = 4^{x} + 9^{x} - 2 . 6^{x} = 6^{x} - 9^{x}$ $\Rightarrow 4^{x} + 2 . 9^{x} - 3 . 6^{x} = 0$ $\sqrt{4^{x} 9^{x}} = 6^{x}$ $\Rightarrow a + 2 b - 3 \sqrt{a b} = 0, a, b > 0$ $\Rightarrow {(a + 2 b)}^{2} = 9 a b$ $\Rightarrow a^{2} + 4 b^{2} + 4 a b = 9 a b$ $\Rightarrow a^{2} + 4 b^{2} - 5 a b = 0$ $\Rightarrow 4 {(\frac{b}{a})}^{2} - 5 (\frac{b}{a}) + 1 = 0$ $\Rightarrow \frac{b}{a} = {(\frac{9}{4})}^{x} = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8} = \frac{1}{4}; 1$ $\Rightarrow x_{1, 2} = {l o g}_{9 / 4} (1 / 4); 0 = \frac{- l n (4)}{- l n (4 / 9)}; 0 =$ $= \frac{l n (4)}{l n (4) - l n (9)}; 0 = \frac{1}{1 - l n (9) / l n (4)}; 0 =$ $= \frac{1}{1 - l n (3) / l n (2)}; 0 = \frac{1}{1 - {l o g}_{2} (3)}; 0$ $\Rightarrow$ $x_{1, 2} = \frac{1}{1 - {l o g}_{2} (3)}; 0 = {l o g}_{9 / 4} (1 / 4); 0$
	Answered by Rasheed.Sindhi last updated on 22/Nov/23
	$(2^x −3^x )^2 =6^x −9^x 4^x +9^x −2∙6^x −6^x +9^x =0 4^x +2(9^x )−3(6^x )=0 (4^x /6^x )+((2(9^x ))/6^x )−3=0 ((2/3))^x +2((3/2))^x −3=0 ((2/3))^x =y y+(2/y)−3=0 y^2 −3y+2=0 (y−1)(y−2)=0 y=1,2 ((2/3))^x =1,2 { ((((2/3))^x =((2/3))^0 ⇒x=0)),((((2/3))^x =2⇒2^(x−1) =3^x )) :} (x−1)log_2 2=xlog_2 3 x−xlog_2 3=1 x=(1/(1−log_2 3 ))$
	${(2^{x} - 3^{x})}^{2} = 6^{x} - 9^{x}$ $4^{x} + 9^{x} - 2 \cdot 6^{x} - 6^{x} + 9^{x} = 0$ $4^{x} + 2 (9^{x}) - 3 (6^{x}) = 0$ $\frac{4^{x}}{6^{x}} + \frac{2 (9^{x})}{6^{x}} - 3 = 0$ ${(\frac{2}{3})}^{x} + 2 {(\frac{3}{2})}^{x} - 3 = 0$ ${(\frac{2}{3})}^{x} = y$ $y + \frac{2}{y} - 3 = 0$ $y^{2} - 3 y + 2 = 0$ $(y - 1) (y - 2) = 0$ $y = 1, 2$ ${(\frac{2}{3})}^{x} = 1, 2$ ${\begin{cases} {(\frac{2}{3})}^{x} = {(\frac{2}{3})}^{0} \Rightarrow x = 0 \\ {(\frac{2}{3})}^{x} = 2 \Rightarrow 2^{x - 1} = 3^{x} \end{cases}$ $(x - 1) \log_{2} 2 = x \log_{2} 3$ $x - x \log_{2} 3 = 1$ $x = \frac{1}{1 - \log_{2} 3}$
	Commented by Bayat last updated on 22/Nov/23
	tnhaks alote
	Commented by Bayat last updated on 22/Nov/23

	Answered by Rasheed.Sindhi last updated on 22/Nov/23

	${(2^{x} - 3^{x})}^{2} = 3^{x} (2^{x} - 3^{x})$ ${(2^{x} - 3^{x})}^{2} - 3^{x} (2^{x} - 3^{x}) = 0$ $(2^{x} - 3^{x}) (2^{x} - 3^{x} - 3^{x}) = 0$ $2^{x} = 3^{x} \lor 2^{x} = 2 (3^{x})$ $x = 0 \lor 2^{x - 1} = 3^{x}$ $(x - 1) \log_{2} 2 = x \log_{2} 3$ $(x - 1) (1) = x \log_{2} 3$ $x = \frac{1}{1 - \log_{2} 3}$
	Answered by witcher3 last updated on 22/Nov/23

	$6^{x} - 9^{x} = (2^{x} - 3^{x}) . 3^{x}$ $\Leftrightarrow (2^{x} - 3^{x}) (2^{x} - 3^{x}) - (2^{x} - 3^{x}) . 3^{x} = 0$ $\Leftrightarrow (2^{x} - 3^{x}) (2^{x} - . 2 . 3^{x}) = 0$ $2^{x} = 3^{x} \Rightarrow x = 0$ $2^{x} - 2 . 3^{x} = 0$ $x = . \frac{\ln (2)}{\ln (2) - \ln (3)}$
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