# 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?

Oct 7, 2016

The equation of tangent is $x - e y = 0$ and slope is $\frac{1}{e}$

#### Explanation:

Let the desired tangent be at $x = {x}_{0}$. As at $x = {x}_{0}$, $y = \ln {x}_{0}$, we are seeking tangent at $\left({x}_{0} , \ln {x}_{0}\right)$

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

the slope of tangent at is $\frac{1}{x}$ and at $x = {x}_{0}$, it is $\frac{1}{x} _ 0$.

Equation of a line passing through point $\left({x}_{1} , {y}_{1}\right)$ and having a slope $m$ is given by

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Hence equation of tangent is $y - \ln {x}_{0} = \frac{1}{x} _ 0 \left(x - {x}_{0}\right)$ or

$y - \ln {x}_{0} = \frac{x}{x} _ 0 - 1$

As the tangent passes through $\left(0 , 0\right)$

$0 - \ln {x}_{0} = \frac{0}{x} _ 0 - 1$ or $- \ln {x}_{0} = - 1$ or ${x}_{0} = e$

and equation of tangent is

$y - \ln e = \frac{x}{e} - 1$ or $y - 1 = \frac{x}{e} - 1$

or $x - e y = 0$ and slope is $\frac{1}{x} _ 0 = \frac{1}{e}$
graph{(y-lnx)(x-ey)=0 [-2.333, 7.667, -1.4, 3.6]}