SOLUTION: 2xlog1 + 2xlog2 - 4xlog3 = 12 xlog1 + xlog2 + xlog3 = 6 3xlog1 + 3xlog2

SOLUTION: 2xlog1 + 2xlog2 - 4xlog3 = 12 xlog1 + xlog2 + xlog3 = 6 3xlog1 + 3xlog2 - 3xlog3 = 3

Question 945621: 2xlog1 + 2xlog2 - 4xlog3 = 12
xlog1 + xlog2 + xlog3 = 6
3xlog1 + 3xlog2 - 3xlog3 = 3

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
solve for x:
2xlog1 + 2xlog2 - 4xlog3 = 12

x=12/(2(log(2)-2log(3))≈-9.1854
..
xlog1 + xlog2 + xlog3 = 6
0+xlog(2)+xlog(3)=6
x(log(2)+log(3))=6
x=6/(log(2)+log(3))≈7.7106
..
3xlog1 + 3xlog2 - 3xlog3 = 3
0 + 3xlog2 - 3xlog3 = 3
3x(log(2)-log(3))=3
x=3/3(log(2)-log(3))≈-5.6789
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