2^(x+y=) 6^y 3^x =3(2^(y−1) ) solve for x and y


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Question Number 58862 by George Mark Samuel last updated on 01/May/19

$2^{x + y =} 6^{y}$ $3^{x} = 3 (2^{y - 1})$ $solve for x and y$
	Answered by MJS last updated on 01/May/19

	$2^{x + y} = 6^{y}$ $2^{x} \times 2^{y} = 2^{y} \times 3^{y}$ $2^{x} = 3^{y}$ $x \ln 2 = y \ln 3$ $y = \frac{\ln 2}{\ln 3} x$ $3^{x} = 3 (2^{y - 1})$ $3^{x - 1} = 2^{y - 1}$ $(x - 1) \ln 3 = (y - 1) \ln 2$ $x \ln 3 - \ln 3 = y \ln 2 - \ln 2$ $y = \frac{\ln 3}{\ln 2} x - \frac{\ln 3}{\ln 2} + 1$ $\frac{\ln 2}{\ln 3} x = \frac{\ln 3}{\ln 2} x - \frac{\ln 3}{\ln 2} + 1$ $\frac{\ln 3}{\ln 2} = a$ $\frac{x}{a} = a x - a + 1$ $x = a^{2} x - a^{2} + a$ $x (1 - a^{2}) = a - a^{2}$ $x (1 - a) (1 + a) = a (1 - a)$ $x = \frac{a}{a + 1} = \frac{\ln 3}{\ln 6}$ $y = \frac{1}{a} x = \frac{1}{a + 1} = \frac{\ln 2}{\ln 6}$
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