Thread: tangent through the origin for lnx

1. 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?

2. 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

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

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

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

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

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

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

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

So the x-coordinate is x=e.

I hope this helps.

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y=lnx find tangent line that passes through the origin

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