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avatar+409 

Solve for the rational numbers x and y:

 

2^(x+y) * 3^(x-y) * 6^(2x+2y) = 72

 

Express your answer as an ordered (x,y) pair .

 Jan 17, 2019
 #1
avatar+533 
+1

6^(2x+2y) is 2^(2x+2y)*3^(2x+2y), so it is now 2^(3x+3y)*3^(3x+y)=72.

 

with some intuition, you can see that 2^3*3^2=72, so 3x+3y=3 and 3x+y=2.

 

now you subtract the second equation from the first so 2y=1 so y=1/2.

 

plugging back in, you can see that 1/2 is also x.

 

so the answer is (1/2, 1/2).

 

HOPE THIS HELPED!

 Jan 17, 2019
 #2
avatar+111357 
+3

2^(x+y) * 3^(x-y) * 6^(2x+2y) = 72

 

2^(x + y) * 3^(x - y) * ( 6^2 )^(x + y)  =  72

 

2^(x + y) * 3^(x - y) * (36)^(x + y) = 72

 

[ 2 * 36]^(x + y) * 3^(x - y) = 72

 

[72]^(x + y) * 3^(x - y) =  72

 

If we let x, y =  1/2 we have that

 

2^1 * 3^0 * 6^2 = 72

 

2 * 1 * 36 = 72

 

So

 

(x, y) =  (1/2, 1/2)

 

cool cool cool

 Jan 17, 2019
 #3
avatar+109468 
+4

2^(x+y) * 3^(x-y) * 6^(2x+2y) = 72

 

\(2^{x+y} * 3^{x-y} * 6^{2x+2y}= 72\\ 2^{x+y} * 3^{x-y} * 2^{2(x+y)}*3^{2(x+y)}= 3^2*2^3\\ 3^{x-y+2x+2y} * 2^{3(x+y)}= 3^2*2^3\\ 3^{3x+y} * 2^{3x+3y}= 3^2*2^3\\ so\\ 3x+y=2\quad(1a)\\ 3x=2-y \quad (1b)\\ 3x+3y=3\quad(2a)\\ x+y=1\quad(2b)\\ y=1-x\quad(2c)\\ sub (2c) into (1b)\\ 3x=2-(1-x)\\ 3x=1+x\\ 2x=1\\ {\bf{x=0.5,\quad y=0.5}} \\ check\\ 2^1*3^0*6^{2}=72\qquad \text{Das ist ausgezeichnet!} \)

.
 Jan 17, 2019

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