Suppose that $f(1) = 1, f(10) = 2, f(100) = 3.$ Let's suppose further that you measure position on your paper in centimeters, with the origin being at the origin of your graph.
If you plot $\log(x)$ vs $f(x)$, you'll plot points at $(0cm, 1cm), (1cm, 2cm),$ and $(2cm, 3cm)$.
If, on the other hand, you use the log paper's log-scale on the x-axis, let's suppose that the first "Decade" of the paper starts at the 0cm mark in the horizontal direction, and the leftmost vertical line of this decade is labelled "1", the second decade starts at the 1cm mark (and starts with "10"), and so on. Then for the point $f(10) = 2$, you'll go to $(1cm, 2cm)$; the other points you plot will be at $(0cm, 1cm)$ and $(2cm, 3cm)$.
In short, you'll draw the same three points.
The horizontal axis may be labelled $\log x$ in the sense that the physical distance from the vertical axis really is (up to a constant) the logarithm of the $x$-value that made you plot a point; the lines on the paper (and their labels "1", "10", "100") are just a way for you to easily exponentiate these distances to get the original value of $x$. In that sense, the paper is "taking the log" for you as you plot things: you have $x = 100$, you look for the vertical line labelled 100, and put a point there...and its distance from the $y$-axis turns out to be 2, which is $\log 100$.
Personally, I don't like it. I tend to label the thin vertical lines 1, 10, 100, etc., and label the axis $x$, but then I'm not an engineer or physicist. Maybe they know something I don't about graphs...
I've added some graphs to make a proper answer of my comments. For the voltage ratio relation you described, the behavior of the function as $ \ f \ $ "tends to infinity" is
$$\frac{U_{out}}{U_{in}} \ = \ \frac{1}{\sqrt{\left(\frac{f}{f_G}\right)^2+1}} \ \ \longrightarrow \ \ \frac{1}{\sqrt{\left(\frac{f}{f_G}\right)^2}} \ \ = \ \ \frac{1}{\left(\frac{f}{f_G}\right)} \ \ = \ \ \frac{f_G}{f} \ \ . $$
So at frequencies "large compared to $ \ f_G \ \ , $ this looks increasingly like a simple inverse relationship between the frequency and the voltage ratio. On a graph, however, this can be rather difficult to read; moreover, a slightly different exponent from $ \ f^{-1} \ $ would be hard to discern. (The graph at left above shows the voltage ratio as a function of frequency with $ \ f_G = 100 \ $ and frequency running out beyond $ \ 25,000 \ \ . $ To its right is a vertical scale enlargement by about a factor of $ \ 30 \ \ $ with frequency out to $ \ 500,000 \ \ . $ The red lines in the first graph indicate the vertical range in the second graph; the red lines in the second graph mark the frequency range in the first one.)
To improve our ability to read the graph to very small voltage ratios, we can apply logarithms to both sides of the above relation to produce a linear equation (mathematicians and some physicists use natural logarithms, but most other disciplines use base-10):
$$ \log_{10} \frac{U_{out}}{U_{in}} \ \ = \ \ \log_{10} \frac{f_G}{f} \ \ \Rightarrow \ \ \log_{10} \frac{U_{out}}{U_{in}} \ \ = \ \ \log_{10} f_G \ - \ \log_{10} f $$ $$ = \ \ (-1)·(\log_{10} f) \ + \ \log_{10} f_G \ \ , $$
which can be rearranged into a "slope-intercept form" with the logarithm of $ \ f \ $ as the independent variable on a line of slope $ \ (-1) \ $ and a $ \ y-$intercept of $ \ \log_{10} f_G \ \ , $ for "sufficiently large" frequencies (as in the graph below, with $ \ f \ $ running from about $ \ 0.001 \ $ to $ \ 400,000 \ ). $ It is possible here to see that beyond a frequency of about $ \ 1000 \ \ , $ the voltage ratio falls by "a decade" for each "decade" increase in frequency.
Any sort of "power-law" function can be treated in the same way to obtain a "log-log" linear equation of the form
$$ \ y \ = \ C·x^n \ \ \longrightarrow \ \ \log_{10} y \ \ = \ \ n· (\log_{10} x) \ + \ \log C \ \ . $$
In experimental practice, data points representing measurements that "follow a power-law" will fall more-or-less along a straight line with a slope that indicates the exponent in the power-law function. (Using natural logarithms instead only changes the numerical values plotted, but has no effect on the observed slope.)
Something similar is done for exponential functions, which also turn up frequently in phenomenological models. There, a "log-linear" transformation is used in which only the ordinate ("vertical coordinate") has a logarithm applied:
$$ y \ = \ C·10^{kx} \ \ \longrightarrow \ \ \log_{10} y \ \ = \ \ \log_{10} [ \ C·10^{kx} \ ] \ \ = \ \ k·x \ + \ \log_{10} C \ \ . $$
Here, there is a straight line in the variable $ \ x \ $ with a slope $ \ k \ \ , $ which gives the "exponential constant" in the relation; the value of this linear function is then $ \ \log_{10} y \ \ . $ (It should be noted that using natural logarithms here will affect the slope, altering it to $ \ (\ln 10)·k \ \ ; $ "natural-logs" would be the proper choice for a model function $ \ C·e^{kx} \ \ . ) $ This last graph shows a full linear and log-linear plot for $ \ 40·10^{-0.02·x} \ \ ; $ the red lines in the graph at right show the extent of the full linear graph.
One last remark that might be made here is that log-log and log-linear plots have "no bottom", since equal steps in the "vertical" direction represent equal factors in $ \ y \ , \ $ so $ \ y = 0 \ $ is transformed to "negative infinity".
[As a historical remark on the "wonderful age in which we live", for quite a long time, one had to purchase specially printed "graph-paper" with "log-log" and "log-linear" grids for these purposes. Once personal computers with graphing utilities became widely available in the 1990's, the market for such items largely evaporated. Not long after that, online graphers also appeared, so it has been quite easy for some while now to make such plots.]
Best Answer
In fact, you have the change of variables : $$\begin{cases} X=\log_{10}(x) \quad;\quad x=10^X \\ Y=\log_{10}(y) \quad;\quad y=10^Y \end{cases}$$ In the new system of axes $(X,Y)$ the equation becomes : $$Y=\log_{10}(m\:10^X+c)$$
FIRST : For $x\to 0$
$X\to -\infty\quad;\quad m10^X\to 0 \quad;\quad Y\to\log_{10}(c)$
$$\text{Horizontal asymptote :}\quad Y=\log_{10}(c)\quad\text{in blue on the figure.}$$
Thus this straight line is related to the constant part of the linear function.
SECOND : For $x\to +\infty \quad;\quad X\to+\infty$
$Y=\log_{10}(m\:10^X+c)=\log_{10}(m\:10^X)+\log_{10}(1+\frac{c}{m\:10^X})$
$Y=\log_{10}(m)+X+\log_{10}(1+\frac{c}{m\:10^X})$
$\frac{c}{m\:10^X}\to 0 \quad;\quad Y\simeq X+\log_{10}(m)$ $$\text{Inclined asymptote}\quad Y= X+\log_{10}(m)\quad\text{in green on the figure}$$ Thus this straight line is related to the slope of the linear function.