# tangent through the origin for lnx

• Apr 6th 2008, 02:47 PM
theowne
tangent through the origin for lnx
Find the value of x where the tangent to y=lnx is a line through the origin.

Can someone give me an idea of the steps involved here? The derivative of lnx will tell the slope, I'm not sure where to go from there. Find a general form of the line equation?
• Apr 6th 2008, 03:09 PM
TheEmptySet
Quote:

Originally Posted by theowne
Find the value of x where the tangent to y=lnx is a line through the origin.

Can someone give me an idea of the steps involved here? The derivative of lnx will tell the slope, I'm not sure where to go from there. Find a general form of the line equation?

Well we know that

$\displaystyle y = \ln(x) \mbox{ and } \frac{dy}{dx}=\frac{1}{x}$

Lets suppose that the x coordinate of the tangent line through the origin is x=a.

Then the ordered pair is $\displaystyle (a, \ln(a))$ and the slope at that point is $\displaystyle m=\frac{1}{a}$

using the above info we can write an equation of the line using the point slope equation.

$\displaystyle y-\ln(a)=\frac{1}{a}(x-a)$ we know that this line passes through the origin or (0,0)

$\displaystyle 0-\ln(a)=\frac{1}{a}(0-a) \iff -\ln(a)=-1 \iff \ln(a)=1 \iff a=e$

Then we know that the point is
$\displaystyle (a,\ln(a)) \iff (e,\ln(e)) \iff (e,1)$

So the x-coordinate is x=e.

I hope this helps.