A line with slope m passes through the origin and is tangent to the graph of y = ln x. What is the value of m? What is the equation of the tangent line?

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Answer 1 · 2016-10-07T15:37:20

The equation of tangent is #x-ey=0# and slope is #1/e#

Let the desired tangent be at #x=x_0#. As at #x=x_0#, #y=lnx_0#, we are seeking tangent at #(x_0,lnx_0)#

Further, slope of tangent to the curve is given by first derivative,

the slope of tangent at is #1/x# and at #x=x_0#, it is #1/x_0#.

Equation of a line passing through point #(x_1,y_1)# and having a slope #m# is given by

#y-y_1=m(x-x_1)#

Hence equation of tangent is #y-lnx_0=1/x_0(x-x_0)# or

#y-lnx_0=x/x_0-1#

As the tangent passes through #(0,0)#

#0-lnx_0=0/x_0-1# or #-lnx_0=-1# or #x_0=e#

and equation of tangent is

#y-lne=x/e-1# or #y-1=x/e-1#

or #x-ey=0# and slope is #1/x_0=1/e#
graph{(y-lnx)(x-ey)=0 [-2.333, 7.667, -1.4, 3.6]}