Recent content by Ericalvusa91

Thank You. I finally got it! Instead of putting the -1 divided by 6, I left it as whole number 1.

I thought I could separate √(1) and √12e^3. Those x-values do not give me the right answer unless I am suppose to do something after that. However, if I already solved for x it should be the answer though right?

I thought you can take the radical one -1-√(1+12e^3)/6 I still get it wrong though

The second value is (-1/3)+√(12e^3)/(6) so should I find the lcd

Heres what i Have 3x^2+x-e^3=0 -b+-√((b^2) - 4ac)/2a -1+-√(1^2)-4(3)(-e^3))/2(3) -1+-√(1+12e^3)/6 Then two possibilities but only need the positive value 1st possibility = -1+1+√(12e^3)/6 = √(12e^3) / 6

Just to be sure c would be -e^3 right?

Nope the positive value is wrong too

I suppose I use the quadratic formula but I'm only suppose to get one answer I get (sqrt (12e^3))/(6) and -(sqrt (12e^3))/(3)

I subtracted the x^-1

So I did what you said. 3x+1 = e^3 x^-1 3x-x^(-1) +1 =e^3 factor out x(3-1^(-1))=e^3 Divide both sides by 3-1^(-1) ans: (e^3)/ 2 But I still get it wrong

Homework Statement ln(3x+1) = 3-lnx Homework Equations Solve for xThe Attempt at a Solution Well I put the ln on the left side ln(3x+1)+ln(x) = 3 Then I combine them ln ((3x+1)(x)) = 3 So I take e e^ln(x(3x+1)) = e^3 I get x(3x+1) = e^3 So I divide the x (3x+1)/(x) = (e^3)/(x)...