Hello, my question is the following:

Question

The derivative of f(x) = ln(x) is given by f'(x) = 1/x. Find an equation of the tangent line to the graph y = ln(x) that passes through the point (0,0).

So if I go through this using the typical method I use I have the point slope formula being y- y1 = m(x - x1).

This gives me y - 0 = m(x - 0) given the values (0,0).

My m is provided which is f'(x) = 1/x.

so i get y - 0 = (1/x)(x-0)

At this point I realize y = (1/x)(x) is what is left over and that is not giving me the correct answer as it would appear to give me 1.

Am I using the wrong method here or am I making a mistake along the way? Any help is appreciated.

Answer 1

the whole problem becomes trivial when you realize that (0,0) does not lie on y = lnx
If you recall your properties and definition of logs, you know that we can only take logs of positive numbers, that is, ln (0) is undefined.

Your method is correct until you come to the point, where the slope would be 1/0 , which of course is undefined.

To make a long story short, the whole question is bogus, since the given point does lie on your curve, thus no tangent is possible.

Answer 2

I think you both misread the question. We want a line tangent to ln(x) which passes through (0,0).

Since the slope of the tangent is 1/x, we want a line through (0,0) and (h,ln(h)) with slope 1/h.

Clearly, at x=e, y=1 and the slope is 1/e, so

y = x/e is the tangent at (e,1)

See

http://www.wolframalpha.com/input/?i=plot+y%3Dln%28x%29%2C+y%3Dx%2Fe+for+x+%3D+0..10

Answer 3

My professor claims that it is solvable because it says that the point (0,0) lieS on a tangent line to the graph y=ln(x) at some point (a,b)

Answer 4

ah thank you steve

Answer 5

why does the line have to cross (h, ln(h))?