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Question Number 136520 by mey3nipaba last updated on 22/Mar/21

Commented by mey3nipaba last updated on 22/Mar/21

$help please .$
	Answered by mindispower last updated on 22/Mar/21
	$⇔ { ((2^x =3^y ⇒x=((yln(3))/(ln(2))))),(((x−1)=((y−1)/(ln(3)))ln(2))) :}$
	$\Leftrightarrow {\begin{cases} 2^{x} = 3^{y} \Rightarrow x = \frac{y l n (3)}{l n (2)} \\ (x - 1) = \frac{y - 1}{l n (3)} l n (2) \end{cases}$
	Answered by Rasheed.Sindhi last updated on 23/Mar/21

	$2^{x + y} = 6^{y} \land 3^{x} = 3 (2^{y - 1})$ $2^{x} = 3^{y} . . . . (i) \land 3^{x - 1} = 2^{y - 1} . . . . (i i)$ $(i) \times (i i) :$ $2^{x + y - 1} = 3^{x + y - 1}$ $\Rightarrow x + y - 1 = 0$ $x + y = 1 . . . . . . . . . . . . . . . . . . (i i i)$ $(i) / (i i) :$ $\frac{2^{x}}{2^{y - 1}} = \frac{3^{y}}{3^{x - 1}}$ $2^{x - y + 1} = 3^{y - x + 1}$ $x - y + 1 = 0 \land y - x + 1 = 0$ $x - y + 1 = y - x + 1$ $x - y = y - x$ $x = y . . . . . . . . . . . . . . . . . . . . . . . . . . (i v)$ $(i i i) & (i v) :$ $x = y = 1 / 2$ ${D o e s n}^{'} t s a t i s f y o r i g i n a l e q u a t i o n .$ $∴ N o s o l u t i o n .$
	Commented by bemath last updated on 23/Mar/21

	$i f x = y = \frac{1}{2} \Rightarrow 2^{x + y} = 2^{\frac{1}{2} + \frac{1}{2}} = 2^{1}$ $R H S \Rightarrow 6^{y} = 6^{\frac{1}{2}} = \sqrt{6}$ $w h y 2 \neq \sqrt{6} s i r ?$
	Commented by Rasheed.Sindhi last updated on 23/Mar/21

	$Y e s m a d a m, p e r h a p s {t h e r e}^{'} s n o s o l u t i o n .$
	Commented by mr W last updated on 23/Mar/21

	$3^{a} = 2^{b} ⇏ a = b = 0$ $i t h i n k {i t}^{'} s w r o n g h e r e :$ $2^{x - y + 1} = 3^{y - x + 1}$ $⇏ x - y + 1 = 0 \land y - x + 1 = 0$ $\Rightarrow x - y + 1 = (y - x + 1) \log_{2} 3$ $\Rightarrow x - y = \frac{\log_{2} 3}{1 + \log_{2} 3}$
	Commented by Rasheed.Sindhi last updated on 23/Mar/21

	$T h α n k s f o r g u i d l i n e s i r!$
	Answered by mr W last updated on 23/Mar/21

	$2^{x} 2^{y} = 2^{y} 3^{y}$ $2^{x} = 3^{y}$ $\Rightarrow x = y \log_{2} 3 = k y$ $w i t h k = \log_{2} 3$ $3^{x - 1} = 2^{y - 1}$ $y - 1 = (x - 1) \log_{2} 3 = (x - 1) k$ $y - 1 = (k y - 1) k$ $\Rightarrow y = \frac{1 - k}{1 - k^{2}} = \frac{1}{1 + k} = \frac{1}{1 + \log_{2} 3} \approx 0.386853$ $\Rightarrow x = \frac{k}{1 + k} = \frac{1}{1 + \frac{1}{k}} = \frac{1}{1 + \log_{3} 2} \approx 0.613147$
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