How to prove units when you have logs in an equation

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Discussion Overview

The discussion revolves around the challenge of proving units in equations involving logarithmic functions, particularly in the context of pressure calculations. Participants explore the implications of using logarithms and the dimensionality of the variables involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how to prove the units when using logarithmic equations, specifically mentioning an equation involving pressure.
  • Another participant notes that logarithmic functions typically yield dimensionless results unless the units are defined in a logarithmic context, such as decibels.
  • There is a discussion about the necessity for the argument of the exponential function to be dimensionless, with one participant expressing confidence that their derived pressure units are correct.
  • A participant challenges the original equation's form and suggests a rearrangement that leads to a different expression, questioning the derivation process.
  • Another participant suggests that a constant may have been lost in the derivation of the formula, implying a potential oversight in the calculations.
  • One participant provides a more complex equation involving simultaneous equations and seeks feedback on their transposition and simplification process.
  • There is a reiteration of the difficulty in performing unit analysis with logarithmic and trigonometric functions, with a reference to the need for dimensionless arguments.
  • A participant expresses uncertainty about how to prove the units in their example, suggesting that constants may play a role in unit cancellation.
  • Another participant agrees with the validity of the formula but expresses uncertainty about its form and implications for calculations.

Areas of Agreement / Disagreement

Participants express a range of views on the validity of the equations and the treatment of logarithmic functions in unit analysis. There is no consensus on how to definitively prove the units in the context provided, and multiple competing perspectives remain.

Contextual Notes

Some participants note that the original equation may have been incorrectly transcribed from memory, and there are discussions about the potential loss of constants that could affect unit consistency.

gtbiyb
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Hi,

What happens when, following a numerical calculation, you have to prove the units but the equation has logs?

For example:

ln P = (ln z1 - ln z2)/(z1 - z2)

So as the RHS will simplify to a number when all values are known you end up with:

ln P = x, x being some value

so to find P you now have to:

e^x = P

I now the units turn out to be Pa as its pressure but how to prove it!

Cheers

Matt
 
Last edited:
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when logs are involved the units of measure would be dimensionless except if the units of measure were based on a log definition like decibels.

A similar example would be using radian measure in trig functions what's the unit of measure of sin(alpha) as in the equation:

(length of hypotenus in meters)*sin(alpha) = (length of opposite side in meters)

sin(alpha) its dimensionless right?
 
Last edited:
Yeah i see that thanks, i have also just read that in e^x the x has to be dimesionless. But my answer should be a pressure in Pa! I transposed a formula to get that above but am confident its valid.
 
Your original equation looks a bit strange to me, where did it come from?
If you rearrange it I get...
lnP = (lnz1 - lnz2)/z1 - z2
i.e lnP = ln(z1/z2)/(z1-z2)
i.e (z1-z2)lnP = ln(z1/z2)
i.e P ^(z1-z2) = z1-z2

Is this what you started with ?
 
Revise the way you have obtained your formula. May be you solved a diff. equation, and a constant has been lost?
 
Thanks guys,

So i have put thought into this in general but in other situations the units cancel as the log is taken of a ratio but in this case i have basically a simultaneous equation:

P(z1)=P(0)e(-z1/λ)
and
P(z2)=P(0)e(-z2/λ)

I know P(z1), P(z2), z1 and z2

I want to know P(0) So i transpose both to eliminate λ thus:

ln P(z1) - ln (P(0)) = -z1/λ

so

λ= -z1/ln P(z1) - ln (P(0))

Therefore:

z1/(ln P(z1) - ln (P(0)))=z2/(ln P(z2) - ln (P(0)))

Simplifying to get:

(z2ln P(z1)-z1ln P(z2)/(z2-z1)=ln (P(0))

What do you think?
 
I can see the original equation i put down was wrong as i was doing it from memory and it was the general theory i was after! The above is correct though.

Matt
 
Anyone got any advice on this?
 
gtbiyb said:
Anyone got any advice on this?

I think we already answered your question. You can't do units math when log and trig funcs are used. The exception being when a measurement is log based such as decibels.

http://en.wikipedia.org/wiki/Decibels
 
  • #10
But what do you do then to prove the units in the example i gave, I'm sure my transposition is correct.
 
  • #11
gtbiyb said:
But what do you do then to prove the units in the example i gave, I'm sure my transposition is correct.

do the units work out when you treat the log funcs as dimensionless?

also you could be missing a constant that provides some additional units for cancelling out

like the G in the classical Gm1m2/r^2 gravity formula the result is the force so the G must have units
that cancel out so the resultant answer is in Newtons.
 
Last edited:
  • #12
Gtbiyb, though ln(P) looks unaccustomed for me, I think you are right, and the formula can be applied for calculations. Both of its parts has the same dimension. If one applies exp to the both of the parts of the formula it will be obtained P0=P(z1)^(z2/(z2-z1))/P(z2)^( (z1/(z2-z1)). I’m not sure it is a better form.
 

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