2-x-y-6-y-3-x-3-2-y-1-solve-for-x-and-y-

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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