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What happens to units in an exponent?

My math textbook just introduced the exponential equation:

$$A_t = Pe^{rt}$$

I've always made it a point in solving math problems to include the units in every calculation.

After I plug in my values:

$$A_{9 years} = $980e^{.07(9 years)}$$

...and simplify:

$$A_{9 years} = $980e^{.63 years}$$

I end up with a unit in my exponent: $e^{.63 years}$

I'm pretty sure this is insolvable, as both Wolfram|Alpha, and Google wouldn't give me an answer. So is this a bad equation?

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    $\begingroup$ Hagen is right. What is $r$? It's probably given as something like "$7\%$ per year", right? Which means that its value actually is $0.07\ \text{year}^{-1}$. $\endgroup$ – Rahul Feb 27 '13 at 19:04
  • $\begingroup$ Google: 7% per year * 9 years $\endgroup$ – Web_Designer Feb 27 '13 at 19:17
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Units in exponents don't make sense. Instead, this hints to the fact that $r$ should have a unit like ${\mathrm s}^{-1}$ so that $rt$ is dimensionless. ($P$ again will carry the unit of whatever this expression calculates in the end).

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  • $\begingroup$ How is $s^{-1}$ a unit? Please explain. $\endgroup$ – Web_Designer Feb 27 '13 at 19:03
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    $\begingroup$ @Web_Designer The second (unit symbol $\mathrm s$) is the SI unit for time, hence $\mathrm s^{-1}$ is the unit e.g. of frequency or whatever "dimensionless per second". Of course $1\,\mathrm s^{-1}=60\,\mathrm{min}^{-1}=3600\,\mathrm h^{-1}=86400\,\mathrm d^{-1}\approx\pi\cdot 10^7\,\mathrm a^{-1}$ are other possible ways to express the same situation per different units. $\endgroup$ – Hagen von Eitzen Feb 28 '13 at 22:07

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