SOLUTION: x^2+y^2=12 x^2+y=10 log[x](2y)=3 log[x](4y)=2

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Question 374726: x^2+y^2=12
x^2+y=10



log[x](2y)=3
log[x](4y)=2
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.x%5E2%2By%5E2=12
2.x%5E2%2By=10
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From eq. 2,
x%5E2=10-y
Substitute into eq. 1,
10-y%2By%5E2=12
y%5E2-y-2=0
%28y-2%29%28y%2B1%29=0
Two solutions,
y-2=0
y=2
Then
x%5E2=10-2
x%5E2=8
x=0+%2B-+2%2Asqrt%282%29
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y%2B1=0
y=-1
Then,
x%5E2=10%2B1
x%5E2=11
x=0+%2B-+sqrt%2811%29
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log%28x%2C%282y%29%29=3 is equivalent to x%5E3=2y
y=%281%2F2%29x%5E3
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log%28x%2C%284y%29%29=2 is equivalent to x%5E2=4y
y=%281%2F4%29x%5E2
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%281%2F2%29x%5E3=%281%2F4%29x%5E2
2x%5E3-x%5E2=0
x%5E2%282x-1%29=0
Two solutions:
x=0
Then
y=0
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2x-1=0
2x=1
x=1%2F2
Then,
y=%281%2F2%29x%5E3
y=%281%2F2%29%281%2F2%29%5E3
y=1%2F16
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