Find the equation of the tangent to the curve defined by y = (ln x) - 1 that is parallel to the straight line with equation 3x - 6y - 1 = 0

The slope of the line should be 0.5 by my calculations. The rest, I have no clue how to do. I do it differently every time. Just one of those days.

well uh.. take the derivative of (ln x) - 1 and set it equal to .5 (i dont know the derivative of ln x rite now, sry)

plug it into (ln x) - 1, find the resulting value, and use the point-slope form of a linear equation.

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well I probably should not be giving you answers to your homework but it is y=1/2x - ln2 -2


first find the slope of the tangent line

if the tangent line is parrallel to 3x - 6y - 1 = 0
then
y= 1/2y - 1/6
so mt is 1/2

now find the deverative
if y = ln x -1
y' = 1/x

now y' = 1/2 when x=2
y= ln 2 -1

therefore the point of tangency is (2, {ln 2 - 1})

y - y1 = m(x - x1)
y - (ln 2 -1) = 1/2(x - 2)
y = 1/2x ln 2 - 2
y ~= 0.5x - 1.30685

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