Question 945621
solve for x:
2xlog1 + 2xlog2 - 4xlog3 = 12
{{{log(1^(2x))+log(2^(2x))-log(3^(4x))=12}}}
{{{2xlog(1)+2xlog(2)-4xlog(3)=12}}}
{{{0+2x(log(2)-2log(3))=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