# Units in exponent - e.g. solve: $2^{3 years}$

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?

• Hagen is right. What is $r$? It's probably given as something like "$7\%$ per year", right? Which means that its value actually is $0.07\ \text{year}^{-1}$.
– user856
Commented Feb 27, 2013 at 19:04
• Google: 7% per year * 9 years Commented Feb 27, 2013 at 19:17

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).

• How is $s^{-1}$ a unit? Please explain. Commented Feb 27, 2013 at 19:03
• @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. Commented Feb 28, 2013 at 22:07