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ln(4x-2) = ln(8x) I keep getting -1/2 but the answer is no solution, I dont get why?
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whats in
\[\Large\rm 4x-2=8x\]Solving for x gives us,\[\Large\rm x=-\frac{1}{2}\]But you need to plug it back into the equation to make sure the solution works.
\[\Large\rm \ln(-4)\]Uh oh! :O
log of negative numbers is not defined in real domain.
oh ok thanks, i get it now
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Good!
completely forgot about negative logs lol
No such thing as negative logs my friend.
there is, but in complex domain
I must study more advanced mathematics then
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