Units of Function

Question

What will be the unit of a function $f(V)=e^V$, where $V$ is in Volts? I mean what will be the unit when the function is written in an implicit format and what will be the units when function has a logarithmic or exponential.

Answer 1

If you're raising $e$ to the power of a physical unit then something has gone wrong—the exponent should be a pure number. So you might have something like $$V=V_0\cdot e^\frac{t}{T}$$

with $\frac tT$ being the ratio of two times.

In electronics that might be

$$V=V_0\cdot e^{-\frac{t}{CR}}$$

where $t$ represents time, $C$ a capacitance and $R$ a resistance—but when you work it out, $CR$ turns out to have the dimension of time: in fact it's the time a capacitor of value $C$ would take to discharge to $\frac1e$ of its original voltage through a resistor of value $R$. So the exponent $-\frac{t}{CR}$ is still just a ratio, not needing physical units. And when you exponentiate it, what comes out is just a numerical value.

The exponent needs to be a pure number since otherwise, exponentiating gives a physically meaningless result.

For example, suppose $x$ represents a length. What does $e^x$ represent? Expanding its power series:

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$

$1$ is a number, $x$ is a length, $\frac{x^2}{2!}$ is an area, $\frac{x^3}{3!}$ is a volume, $\frac{x^4}{4!}$ is a hypervolume . . . The individual terms represent things which it's meaningless to add together. And the only way to avoid this is to make sure that $x$ is a dimensionless quantity, i.e. one that can be expressed just as a number without any physical unit.