Let a > 0 and b > 0. If u and v are any real numbers, then:

1. a^ua^v = a^u+v
2. (a^u)^v = a^uv
3. (ab)^u = a^ub^u
4. a^u/a^v = a^u-v
5. (a/b)^u = (a^u/b^u)
6. (d/dx)(a^u) = (a^uln a)(du/dx)
7. ∫a^udu = (1/ln a)a^u + C

Proof:
Using the definition a^x = e^xlna
and the theorems:
e^pe^q = e^p+q,
ln pq = ln p + ln q, and
ln p/q = ln p - ln q

1. 
   a^ua^v
   = e^{uln a}e^{vln a}
   = e^{uln a +vln a}
   = e^{(u+v)ln a}
   = a^u+v
2. 
   (a^u)^v
   = e^{uvln a}
   = a^uv
3. 
   (ab)^u
   = e^{uln (ab)}
   = e^{u(ln a + ln b)}
   = e^{uln a + uln b}
   = e^{uln a}e^{uln b}
   = a^ub^u
4. 
   a^u/a^v
   = e^{uln a}/e^{vln a}
   = e^{uln a}e^{-vln a}
   = e^{uln a - vln a}
   = e^{(u-v)ln a}
   = a^u-v
5. 
   (a/b)^u
   = e^{uln (a/b)}
   = e^{u(ln a - ln b)}
   = e^{uln a - uln b}
   = e^{uln a}e^{-uln b}
   = e^{uln a}/e^{uln b}
   = a^u/b^u
6. (d/dx)(a^u)
   = (d/dx)e^{uln a}
   = e^{uln a}(d/dx)(uln a)
   = e^{uln a}ln a(du/dx)
   = (a^uln a)(du/dx)
7. Since F'(x) = f(x), working backwards:
   (d/dx)(1/ln a)a^u + C
   = (1/ln a)(a^uln a)(du/dx)
   = a^u(du/dx)
   so, ∫a^udu = (1/ln a)a^u + C

These laws can be useful in reducing exponents which can make exercises easier. Also, some multiple choice standardized tests might have the answers listed in a way that is different than your answer. It may be possible to change your answer using these laws to match one of the answers they provide.