Equation for Tangent Line that passes through $(0,1)$ on the curve $y = \ln x$

Question

I'm totally lost. I've been trying to figure this out. This is what I've figured out:

$dy/dx = 1/x$

$y$-intercept $= 1$

So I try to do $y-y_1 = m(x-x_1)+b,$ which I get as $y-1 = 1/x(x-0)+1,$ simplified to $y = 3.$

But I feel like that is totally wrong and well, obviously it isn't even an equation really.

Can someone help me out with this?

Answer 1

Let $(x,\ln x)$ be the point of tangency.

Then the slope of the tangent line is given by $m=\frac{1}{x}$, as you have, and the slope is also given by

$\displaystyle m=\frac{\ln x-1}{x-0}=\frac{\ln x-1}{x}$.

Setting these two expressions equal gives $\ln x-1=1$, so $\ln x=2$, $x=e^2$, and so $m=\frac{1}{e^2}=e^{-2}$.

Therefore the tangent line has equation $y=\frac{1}{e^2}x+1$.

Answer 2

let us pick a general point $(a, \ln a)$ on the graph of $y = \ln x.$ the tangent line at this point has slope $\dfrac1a.$ the equation of the tangent line therefore is $$y - \ln a = \dfrac1{a}(x-a)$$ suppose this linen goes through $(0,1).$ that requires $$1-\ln a=-1 $$ that is $$a = e^2.$$ the tangent at the point $(e^2, 2)$ will go through $(0,1).$