The curve is given as y = ln(x) - 1. Find the equation of the tangent to this curve that is parallel to the line 3x-6y-1 = 0.

Ok, so i differientiated y = ln(x) - 1 and found x = -1/2

Something is not right here...

plz helpp

Question

The curve is given as y = ln(x) - 1. Find the equation of the tangent to this curve that is parallel to the line 3x-6y-1 = 0.

Ok, so i differientiated y = ln(x) - 1 and found x = -1/2

Something is not right here... 

plz helpp

anonymous · Answer

$$\frac{1}{x}=\frac{1}{2}$$ solve for $x$ in your head

anonymous · Answer

x = 2?

anonymous · Answer

yes

anonymous · Answer

slope is $\frac{1}{2}$, point is $(2,\ln(2)-1)$

anonymous · Answer

what? but i got x = -1/2...

the derivative is dy/dx = 1/x and i set dy/dx = -2 and solved for x

anonymous · Answer

cause i thought the slope was parallel to the other line

anonymous · Answer

$$3x-6y-1 = 0$$ what is the slopoe?

anonymous · Answer

oh, nvm, i see. I thought it was perpendicular

anonymous · Answer

k

anonymous · Answer

thank you !

anonymous · Answer

yw