How to calculate the tangent line of $\ln(x)$ through $(2,6)$ by hand?

Question

I came up with my own question for a tutor student, but now I am stuck myself.

The exercise was to calculate the equation of a tangent-line of $f(x) = \ln(x)$, which goes through the point $(2,6)$. I'm trying to solve the problem by hand, but the equations I get to solve the problem aren't solveable with the lineair algebra I'm aware of. Is this right, or am I missing something?

Representation of the problem, where the green line represents $\ln(x), A = (2,6)$ and the black line is the unknown which should be calculated:

Answer 1

You know that the inclination of the line is equal to the derivative of the function on the point in witch they intersect. Then

$ \cfrac{6-\ln x}{2-x}=\cfrac{1}{x}$ which implies

$x(7-\ln(x))- 2 = 0$.

Can you solve from here?

Answer 2

Derivative of $ \log x$ is $\dfrac{1}{x}$.

Point/slope form of a straight line:

$$ \dfrac{log\, x -6}{x-2}= \frac{1}{x}$$

Simplify

$$ x \,log\, x -7x +2 =0 $$

Only numerical solution possible to find tangent point from this transcendental equation. Has two solutions with approx. values:

$$x=(0.23697, 1094.6 ) $$

Take logs of above to find $y$ values for the two $x$ values.

$$y= (-1.43981, 6.99817).$$

EDIT1:

In your sketch only the first root appears correctly. In order that the second root be also seen, a separate graph has to be made with a larger domain upto say $x=1100$ and aspect ratio say $\dfrac12.$

Answer 3

The equation $$x \log( x) -7x +2 =0$$ has two solutions in terms of Lambert function. They are $$x_1=-\frac{2}{W_{-1}\left(-\frac{2}{e^7}\right)}\approx0.236972 \qquad \text{and} \qquad x_2=-\frac{2}{W\left(-\frac{2}{e^7}\right)}\approx1094.63$$

The bounds for the smallest root can be estimated using $$-1-\sqrt{2u}-u < W_{-1}(-e^{-u-1}) < -1-\sqrt{2u}-\frac{2}{3}u$$ Using $u=6-\log(2)$, this gives $$-7+\log (2)-\sqrt{2 (6-\log (2))}<W_{-1}\left(-\frac{2}{e^7}\right)<-5+\frac{2 \log (2)}{3}-\sqrt{2 (6-\log (2))}$$ that is to say $$0.209102 < x_1 <0.256549$$ Starting at $x_0=\frac e{10}$ (this is close to the midpoint), the first iterate of Newton method will be $$x_1=\frac{20-e}{10 (5+\log (10))}\approx 0.236652$$