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 $y=3x+2$.
 A: The slope of the sought tangent line to the curve $f(x)=\ln{x}$ is going to be parallel to the line $y=2x-3$ which means that their slopes should be the same. Now, find what x-value (let's call it $x_0$) for the curve $f(x)=\ln{x}$ corresponds to a slope of $2$:
$$\frac{1}{x_0}=2\implies x_0=\frac{1}{2}$$
We have the slope, the x-value and the last piece of the puzzle that's missing is the y-value (let's call it $y_0$). A point and a slope—that's all the information you need to uniquely determine a line. The tangent line is going through our curve $f(x)=\ln{x}$ at the point $x_0=\frac{1}{2}$:
$$y_0=f\left(\frac{1}{2}\right)=\ln{\left(\frac{1}{2}\right)}=-\ln{2}$$
Now, just use the formula for the equation of a tangent line:
$$y=f'(x_0)(x-x_0)+y_0\\
y=2\left(x-\frac{1}{2}\right)-\ln{2}\implies\\
y=2x-1-\ln{2}.
$$
And that's the line which is tangent to the curve $f(x)=\ln{x}$ and parallel to the line $y=2x-3$ at the same time.
A: The slope of the line y = 2x - 4 = 0 is 2.
The slope of the curve f at the point (x,f(x)) is f'(x).
Solve f'(x) = 2 for x.  Let the solution be a.  
The equation of the line tangent to f and the given line is
y - f(a) = 2(x - a).  
When f = ln, f'(x) = 1/x = 2, x = 1/2.
Where is there a problem?
A: 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
$
A: Any line having the equation $y=mx+c$ , has a slope $m$.

Slope is nothing but the tangent of the angle that the line makes with the X Axis. This angle is measured in the counterclock wise sense. 
$$m=\tan \theta$$
Now when are two lines parallel? When they maintain a fixed distance throughout and never intersect. In other words the slopes of both the lines should be same. 
Both the equations should have the same slope but they can have different intercepts.

Now, the first derivative of any function $y=f(x)$ at any point $(x_0,y_0)$ , actually gives you the slope of a line that is tangent to $f(x)$ at $(x_0,y_0)$.

The slope of the line satisfies:
$$\frac{dy}{dx} \mid_{(x_0,y_0)}=m=\tan \theta$$
Now coming to your question, lets find the general slope of any line that is a tangent to $y=f(x)=\ln x$
$$m=\frac{dy}{dx}\mid_{(x_0,y_0)}=\frac{1}{x_0}$$
Now we find the equation of a general line that passes through $(x_0,y_0)$ with the above slope $m$:
$$y-y_0=\frac{1}{x_0}(x-x_0)$$
since, $(x_0,y_0)$ lies on the curve $y=\ln x$ we have:
$$y_0=\ln x_0$$
So the equation of line becomes:
$$y-\ln x_0=\frac{1}{x_0}(x-x_0) $$
$$y=\frac{x}{x_0}+\ln x_0 -1$$
Now we need this line to be parallel to the line 
$$p:2x-y-3=0$$
$$p:y=2x-3$$
Clearly the slope is the coefficient of $x$ i.e. $m=2$.
In our  original equation if we compare then the slope $m=\frac{1}{x_0}=2$, giving $x_0=\frac{1}{2}$,
Putting this back to the equation we get:
$$y=2x+\ln \frac{1}{2} -1$$
Or
$$y=2x-\ln 2 -1$$
Hope this helps...
