Unit of Measure of Exponentiated Item

Steve Zissou

Hello.
Let's say we have the quantity
f=1/(1+x)
where x has no unit of measure. What is the unit of measure of f, once we take f^t, where t can be in years?
Thanks

tiny-tim

Hello Steve! Welcome to PF!

Still no units.

(f^t = e^tlnf = ∑(tlnf)ⁿ/n!)

Steve Zissou

Hi tiny-tim. Thanks for the reply.
Just a quick thought, if we say
f^t=exp[t ln f]
then we still have that t is years, and exp[time] can't be ok.
Am I right?
Thanks for your help

tiny-tim

ah, it would have to be f^t/t_o

Khashishi

It's not valid to take f^t, with t in years. If f is dimensionless, the exponent has to also be a dimensionless number.

Steve Zissou

Thanks Khashishi.
Can I ask, if the exponent is not dimensionless (as tiny-tim suggested above, by saying it should be t/t0) then does it mean that f must have units that I didn't know about or expect?

Rather, let me ask this: what units would f have, if the exponent has units of time?

Thanks guys

Steve Zissou

Let's generalize. We have the quantity f. Let's say f is distance, so it is in units of meters.
Taking f^2 would give square meters.
Buy let's take it to an exponent that has units, like time.
f^t is now in what units?

Travis_King

the t in that equation should be dimensionless. So, either simply call it the "number" of seconds (or minutes, or years, whatever) or raise f to something like:

f^(t/[1 sec]) to yield a dimensionless number in the exponent.

Good thread here:

Steve Zissou

Travis_King

Thanks

Khashishi

Usually, you have a time expression like:
[itex]Y=A \exp(-t/\tau)[/itex]
where tau is a time constant with the same units as t, so the argument to exp is dimensionless.

Mathematically, you can absorb the time constant into the base of the exponent since
[itex]A \exp(-t/\tau) = \exp(1/\tau)^{-t} = f^{-t}[/itex]
[itex]f=\exp(1/\tau)[/itex]
So, f needs to have units of [itex]\exp(1/years)[/itex] to match up with t in years. No one in their right mind would do something like this, but it makes mathematical sense.

sophiecentaur

There's no reason to expect that you can use a quantity as an exponent. After all, you only need to say, in words, what "the exponent" means. It means the number of times that a number is multiplied by itself and it would be daft to say "Mutiply 3 by itself five point three inches times". Go back to basics for the answers to this sort of question.

Khashishi

This is only one definition, and rather elementary and limited. A lot of natural phenomena exhibit exponential growth or decay. It's probably better to view an exponential as a function whose derivative is proportional to itself.

sophiecentaur

I disagree entirely (and most humbly). Exponential growth is exactly what happens when a fractional increase is repeated a number of times.
Your more sophisticated version is very useful but it's only describing a consequence of the process.

truesearch

the best way to view exponential (natural) growth/decay is to say 'rate of change is proportional to how much you have got'.....this is where I start with students and they seem to be able to relate it to money and savings and interest rates as well as physical phenomena such as radioactive decay
i.e dA/dt = +/-constant x A

This is exactly the same as saying that you get the same fractional increase or decrease
per unit time.

truesearch

The exponent does not have units/dimensions.
It is the powerthat a number (e) is raised to.... just a number.
In the same way a log has no units/dimensions.... it is just a number

cepheid

But your conceptual definition only makes sense for integer exponents. You have to introduce more advanced concepts like the idea of a limit to deal with the more general case anyway, am I right?

sophiecentaur

Mine is a simple, starting definition, true, but it extends, without too much imagination, to non-integers. And, as far as the original question goes, it establishes a logical reason why the index is dimensionless. The logic doesn't change.