haruspex said:
You don't need to assign units to either side of the f(x)= equation, they're dimensionless. But you can rewrite the ##e^x## as ##e^{\lambda x}## where ##\lambda=1m^{-1}##.
In relation to issue of whether ##e^x## is an "error" in an equation describing a physical process when ##x## has a dimension:
In the first place, I don't see arguments of the form "You can rewrite ..." as having any bearing on question. Yes, an equation representing a physical process can be transformed to an equation in dimensionless form, but that doesn't show the original form of the equation is invalid.
Perhaps your complete thought is "Your original equation is wrong or meaningless and you
should rewrite it as ...##.\ ## In the example, I don't see that the original equation is wrong or meaningless in the sense of being uninterpretable or so ambiguous that a person doing in measurements in cm instead of meters couldn't figure out how to rewrite it as different equation ##g(y)## where ##y## has units of cm.
My equation may be wrong in the sense that the task of defining ##g(y)## can't be accomplished by the straightforward use of conversion factors. That's a topic we should investigate!
Let's pursue your suggestion of stating the equation as ##p = f(x) = C( 2 - e^{\lambda x} ) ## where ##\lambda## has units of ##m^{-1}## and ##x## has units of ##m##. Can we convert that equation to a formula ##p = g(y) ## where ##y## has units of cm by using conversion factors?
To convert to cm, we must convert both ##\lambda## and ##x## using the conversion factor (m/100 cm). We have ##100 y \ (cm) = x\ (m) ## and ##\lambda\ (m^{-1}) = \lambda\ (100\ cm)^{-1})## So the equation converts to ##p = g(y) = C(2 - e^{ \frac{\lambda}{100}100 y}) = C(2 - e^y)## But the correct equation (for ##y## in cm) should be something like ##p = g(y) = C(2 - e^{\frac{y}{100}}) ##.
I said "something like" that because we must change the value of ##C## from ##\int_{0}^{ln(2)} {(2 - e^x)} dx ## to ##C_2 = \int_0^{\ln(200)} { ( 2 - e^{\frac{y}{100}} ) } dy ## in order to normalize the probability distribution. We also must convert the interval on which the equation applies from ##[0, ln(2)]## to ##[0, ln(200)]##.
Are we opposed to letting the function ##ln(.)## have an argument with a dimension? If so, how can we justify converting ##ln(2)## to ##ln(200)## ? A dimensionless constant like "C" or "2" can be converted to a different numerical value if it depends on several different dimensions. For example the "1" in F = (1)MA can convert to a different constant if we don't use MKS units. However, the only dimension that has been mentioned in this problem is length [L]. I don't see any way that a dimensionless constant that is define only in terms of lengths can be converted to a different numerical value by changing the unit of measure for length.
In contrast to the above difficulties if we take the viewpoint that the ##x## in ##e^x## and the "2" in ##\ln(2)## have dimension [L] length given in meters then the conversion from meters to cm gives results we need, namely ##e^{\frac{y}{100}}## and ##\ln(200)##.From my point of view the probability density function ##f(x)## is not dimensionless. Like a linear density function for the density of physical mass, it represents "per unit length", so in my equation ##f(x)## has units of (1/meter). However, that consideration still leaves length as the only dimension represented in the equation.