# Dimensionally impossible equation

1. Oct 23, 2015

### Granger

I was reading some stuff about dimensionally impossible equations. It was said that the equation

v = e^bt (b is a constant so that bt is dimensionless)

was dimensionally impossible. I understand that BUT I don't understand what they said next about the dimension of the right-side. They said it was dimensionless, but I thought it had dimension T (like the left-side has dimension LT^-1)... Can someone explain me why I'm wrong? Thanks :)

2. Oct 23, 2015

### jbriggs444

If that equation were written as v(t) = v0ebt then it would make more sense.

For that to make sense, ebt has to be dimensionless.

3. Oct 23, 2015

### Khashishi

You said that $bt$ was dimensionless, so of course $e^{bt}$ is dimensionless also. But this is inconsistent with $v(t)$ having dimensions of length/time, so it is a "dimensionally impossible equation".

Generally speaking, transcendental functions like $e^x$, $\sin(x)$, $\log(x)$, etc. can never operate on units and are always dimensionless. A small exception is when the inverse function appears in the same equation, like $f(t) = e^{4 \log t}$. But it's unclear why you would ever write it this way rather than $f(t)=t^4$.

Last edited: Oct 23, 2015
4. Oct 24, 2015

### Chandra Prayaga

Any such equation has a constant in front, in this case, the constant is 1, in appropriate units. for example, v = e^bt would mean: v = (1 m/s) e^bt, a particular case of the equation written by jbriggs 444 above

5. Oct 24, 2015

### HallsofIvy

If b has dimensions of "1 over time", say, sec^{-1}, then bt is "dimensionless". Then $v(t)= x_0e^{vt}$ has whatever units $x_0$ does.

6. Oct 24, 2015

### Khashishi

No, unless the text specifically states that a particular system of natural units are being used, such as atomic units. You would never see an equation like this with implied SI units.