<s>[INST] find equation of tangent for function $f(x)$ parallel to a line I dont have a clue how to solve this kind of problem.
I dont know how to find out slope of $\ln x$.
$f(x) = \ln x$
line $p: 2x-y-3=0$
I have searched for slope of $\ln x$, I know how find out slope for one point, for example:
first I need to derivate the $\ln x$ ==> $1/x$ then I can calculate slope for every point, for example $x=5$ then it is $1/5$.
Also I know how to find out tangent parallel to $y =x^3$ if line p is $3x-y-4=0$

first step: derivate $y =x^3$ ==> $3x^2$, $y=3x-4$ ==> slope for line is $m=3$ and slope for $y =x^3$ is $3$
second step: $3x^2$=$3$ ==> $x=+-1$

third step: use $y - y0 = m*(x-x0)$
for positive one its $y-1=3*(x-1)$ because $f(1)=1^3$ ==>$ y = 3x-2$ then for negative one its $y=3x+2$.
 [/INST] Equation of tangent to $y = f(x) = \ln(x)$ at $(x_1, y_1)$ is of form
$ (y - y_1) = f'(x_1)(x - x_1) $ where
$ f'(x) = \frac{d}{dx} \ln(x) = \frac{1}{x} $
$
\Rightarrow (y - y_1) = \frac{1}{x_1}(x - x_1)
$
Since Slope of $p: 2x-y-3=0$ is 2, hence,
Slope of || tangent line = Slope of $p$ = 2
$
\Rightarrow \frac{1}{x_1} = 2
\Rightarrow x_1 = \frac{1}{2}
\Rightarrow y_1 = \ln(x_1) = \ln(\frac{1}{2}) = -\ln(2)
$
$ \therefore \ $ Tangent of $f(x)$ || p is:
$
y - (-\ln(2)) = 2(x - \frac{1}{2})
\Rightarrow 2x - y - \ln(2) - 1 = 0
$
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