Let a > 0 and b > 0. If u and v are any real numbers, then:
 a^{u}a^{v} = a^{u+v}
 (a^{u})^{v} = a^{uv}
 (ab)^{u} = a^{u}b^{u}
 a^{u}/a^{v} = a^{uv}
 (a/b)^{u} = (a^{u}/b^{u})
 (d/dx)(a^{u}) = (a^{u}ln a)(du/dx)
 ∫a^{u}du = (1/ln a)a^{u} + C
Proof:
Using the definition a^{x} = e^{xlna}
and the theorems:
e^{p}e^{q} = e^{p+q},
ln pq = ln p + ln q, and
ln p/q = ln p  ln q
a^{u}a^{v}
= e^{uln a}e^{vln a}
= e^{uln a +vln a}
= e^{(u+v)ln a}
= a^{u+v}
(a^{u})^{v}
= e^{uvln a}
= a^{uv}
(ab)^{u}
= e^{uln (ab)}
= e^{u(ln a + ln b)}
= e^{uln a + uln b}
= e^{uln a}e^{uln b}
= a^{u}b^{u}
a^{u}/a^{v}
= e^{uln a}/e^{vln a}
= e^{uln a}e^{vln a}
= e^{uln a  vln a}
= e^{(uv)ln a}
= a^{uv}
(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}
 (d/dx)(a^{u})
= (d/dx)e^{uln a}
= e^{uln a}(d/dx)(uln a)
= e^{uln a}ln a(du/dx)
= (a^{u}ln a)(du/dx)

Since F'(x) = f(x), working backwards:
(d/dx)(1/ln a)a^{u} + C
= (1/ln a)(a^{u}ln a)(du/dx)
= a^{u}(du/dx)
so, ∫a^{u}du = (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.