What is the equation to a line having a slope = 1 that is tangent to the equation y=0.2664Ln(x) + 1.1858?

Answer 1

First you have to find out where this line has a slope of 1 y differentiating: dy/dx=.2664/x=1 when x=.2664. When x=.2664, y=.2664ln(.2664)+1.1858=.8334. Now just use the point slope formula for the rest: y-.8334=1(x-.2664)
or y=x+.567

Answer 2

ANSWER : y = x + 0.567

1) Take the first derivative of the given
equation. y' = 0.2664/x
2) Set y' = 1 , and solve for x = 0.2664 = x1
3 Substitute x from (2) for y from the given
equation. y = 0.8334 = y1
4) Using the point-slope form of a line
y - y1
m = ---------
x - x1
Substitute x1 and y1 and m = 1, simplify
to find the answer above.