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This question already has an answer here:

Is power of only a number or an exponential function is dimensionless? If power of any other thing can also be dimensionless then please explain with examples.

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marked as duplicate by Kyle Kanos, Brandon Enright, jinawee, DavePhD, BMS Apr 11 '14 at 5:35

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – user80551 Apr 10 '14 at 7:41
  • $\begingroup$ if i had understood that, i would have never posted this question. i want an answer which can easily be understood.@user80551 $\endgroup$ – elle Apr 10 '14 at 7:48
  • $\begingroup$ One way to think about it is this: Mathematical identities for functions like the exponential function are about numbers, so in order to sensibly use these functions we must use numbers only as their input. $\endgroup$ – Danu Apr 10 '14 at 8:10
  • $\begingroup$ Related (dupe?) physics.stackexchange.com/q/7668 Partially related physics.stackexchange.com/q/13060 $\endgroup$ – user80551 Apr 10 '14 at 8:55
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    $\begingroup$ Another way of seeing clearly why an exponential's argument should be dimensionless is to Taylor expand: $\exp(x) = 1 +x + x^2 /2 + ...$ Every term has a different dimension if $x$ is dimensionful, and hence cannot be summed. See: damtp.cam.ac.uk/user/tong/relativity/three.pdf for detailed notes on dimensional analysis. $\endgroup$ – JamalS Apr 10 '14 at 9:53
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To consider an example, take the case of exponential decay

$$N=N_\circ e^{-\lambda t}$$

We can write this as

\begin{eqnarray*} N & = & \frac{N_{\circ}}{e^{\lambda t}}\\ & = & \frac{N_{\circ}}{\underbrace{e\times e\times e\times e\times\ldots \times e}_{\lambda t\text{ times}}} \end{eqnarray*}

So $\lambda t$ must be a dimensionless term that is telling how many times we should multiply $e$ by itself. Thus, $\lambda t$ must be dimensionless "overall". Individually, $\lambda$ has the dimensions of $[T^{-1}]$ which cancels with $t$ to give a net dimensionless quantity.

$\underbrace{e\times e\times e\times \ldots}_{10 \text{ meters times}}$ makes no sense mathematically.

We could have taken a dimensional quantity instead of $e$ but the exponent $\lambda t$ would still be dimensionless. eg in the kinematical equation $s=ut + \frac 12 at^2$, $t^2$ has the dimensions of $[T^2]$ but the exponent $2$ is dimensionless.

The same applies to transcendental functions i.e. logarithmic, trigonometric, etc.

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  • $\begingroup$ it is getting impossible for me to bring the alphabets in superscript $\endgroup$ – elle Apr 10 '14 at 10:26
  • $\begingroup$ @elle It's mathjax, see meta.math.stackexchange.com/questions/5020/… $\endgroup$ – user80551 Apr 10 '14 at 10:28
  • $\begingroup$ In $e^{\lambda t}$ ‘λt’ is dimensionless, so we say that ‘$e^{\lambda t}$ ’ (whole) is dimensionless. Now in $e^{\lambda t}$ ‘λt’ is dimensionless, so would $e^{\lambda t}$ (whole) be dimensionless? $\endgroup$ – elle Apr 10 '14 at 10:35
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    $\begingroup$ @elle The exponent ($\lambda t$) is dimensionless but if the base ($e$) is not dimensionless, then as a whole, $e^{\lambda t}$ won't be dimensionless (unless the exponent is zero of course.) Eg $\text {meter}^2$ has the dimensions $[L^2]$. Here, the exponent $2$ is dimensionless. $\endgroup$ – user80551 Apr 10 '14 at 10:42

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