Since you will use several values of $x$, it is probably easiest to find a function for the slope, in terms of $x$, and then plug in the various values for $x$. That is, the slope of the secant line $PQ$ is the rise over run (change in $y$ over change in $x$):
$$m(x) = \frac{x^2 + x + 4 - 24}{x - 4}$$
So, $m(x)$ gives the slope for any particular value of $x$. A practical reason to do this is, for example on a TI-83 or TI-84 or something like it, you can now type in that function to $Y_1$ and then go to the Table and plug in the various $x$ values and you get the slopes immediately. And, since it is so fast, you can check this for more values than the question even asks for to get an even better intuition. Or, you could use Wolfram Alpha to accomplish the same thing. For example, type in:
Evaluate (x^2 + x + 4 - 24)/(x - 4) at x = 4.1, 4.01, 4.001, 3.9, 3.99, 3.999
Back to the problem,
$\begin{align*}
m(4.1) =& \frac{4.1^2 + 4.1 + 4 - 24}{4.1 - 4} = \frac{0.91}{0.1} = 9.1 \\
m(4.01) =& \frac{4.01^2 + 4.01 + 4 - 24}{4.01 - 4} = \frac{0.0901}{0.01} = 9.01 \\
m(4.001) =& \frac{4.001^2 + 4.001 + 4 - 24}{4.001 - 4} = \frac{0.009001}{0.001} = 9.001 \\
m(3.9) =& = \frac{3.9^2 + 3.9 + 4 - 24}{3.9 - 4} = \frac{-0.89}{-0.1} = 8.9 \\
m(3.99) =& = \frac{3.99^2 + 3.99 + 4 - 24}{3.99 - 4} = \frac{-0.0899}{-0.01} = 8.99 \\
m(3.999) =& = \frac{3.999^2 + 3.999 + 4 - 24}{3.999 - 4} = \frac{-0.008999}{-0.001} = 8.999
\end{align*}$
From this, we would probably guess that the slope of the tangent line when $x = 4$ is 9. This does not guarantee that we are right, but assuming the function is reasonably well behaved, we could be pretty confident in this guess.
And, once you learn how to calculate derivatives, you will find this is correct as the slope of the tangent line at any $x$ value is the derivative. Since
$$y'(x) = 2x + 1$$
we see that
$$y'(4) = 9$$
To find the slope of the curve, or the gradient you can differentiate the function and plug in the $x$ value of the point into the derivative and that will yield the gradient at that point. So you have $y=5-x^2$ at the point $(1, 4)$.
$$y'=-2x$$
The gradient of the curve, $y=5-x^2$ at $(1,4)$ is:
$$y'(1)=-2(1)=-2$$
For the equation of the tangent line, you have the gradient $(-2)$, a point on the line $(1, 4)$, and you know that the equation of a line is $y=mx+c$. You can plug in the values that you know to find the value of $c$ and then you'll have the equation of the tangent line.
Best Answer
Once you have found the slope of the line tangent to the curve at a given point $(x_1, y_1)$, you can simply use the point-slope form of a line $y-y_1=m(x-x_1)$ for the equation of your tangent line at this point. In your case, $y-ln(2)=\frac{1}{2}(x-2)$.