# Line tangent to the natural log

I want to find a line that passes through $(0,-1)$ and is tangent to $\ln(x)$.

I've tried saying: ''I want to find a line that has the slope $1/x$ and passes through $(0,-1)$" but this yields:

$$y=kx+m \\ \Rightarrow 1= \frac{1}{0} +m$$

And the logic is already broken at this point.

• Sorry, I meant (0,-1) – user3200098 Apr 13 '14 at 20:08
• Your flaw is plugging in $x=0$ to get the slope. The derivative gives the slope of the tangent for the value of $x$ at the point of tangency, not just anywhere. The line isn't tangent to the curve at $x=0$--in fact, $x=0$ isn't in the domain of the function being graphed. – MPW Apr 13 '14 at 20:14

The line tangent to $\ln{x}$ at $x=a$ is

$$y=\ln{a}+\frac{1}{a}(x-a)$$

You want $y=-1$ when $x=0$, so solve for a in

$$-1=\ln{a}+\frac{1}{a}(-a)$$

• Why is the line given by that equation...? – user3200098 Apr 13 '14 at 20:14
• If you are familiar with Taylor's theorem, then this is just the first two terms of the series. If not, then write it out in point slope form and solve for $y=f(x)$: $$y-y_0=m(x-x_0)$$ $$f(x)-f(a)=f'(a)(x-a)$$ – solstafir Apr 13 '14 at 20:15

Slope between $(0,-1) , ( x, \ln x )$ is
$\dfrac {ln\ x +1}{x}$
$\dfrac {ln\ x +1}{x} = \dfrac {1/x}{1}$