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ln(x^2-3x-1)-ln(x+4)=0 solve for x
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ln[(x^2-3x-1)/(x+3)]=0 (x^2-3x-1)/(x+3)=1
that 3 should be a four lol down below
x^2-3x-1=x+4 x^2-4x-5=0 (x-5)(x+1)=0 x=5,-1
lol k
how did you get one on the other side of the equal sign?
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ln(x^2-3x-1)=ln(x+4) hence i got...
e^blah=e^blah but if the blah is 0 then e^0=1
is that too blah for you?
ohhh ok
I got it. :) thanks
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cool!
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