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

1. auav = au+v
2. (au)v = auv
3. (ab)u = aubu
4. au/av = au-v
5. (a/b)u = (au/bu)
6. (d/dx)(au) = (auln a)(du/dx)
7. ∫audu = (1/ln a)au + C

Proof:
Using the definition ax = exlna
and the theorems:
epeq = ep+q,
ln pq = ln p + ln q, and
ln p/q = ln p - ln q

1. auav
= euln aevln a
= euln a +vln a
= e(u+v)ln a
= au+v

2. (au)v
= euvln a
= auv

3. (ab)u
= euln (ab)
= eu(ln a + ln b)
= euln a + uln b
= euln aeuln b
= aubu

4. au/av
= euln a/evln a
= euln ae-vln a
= euln a - vln a
= e(u-v)ln a
= au-v

5. (a/b)u
= euln (a/b)
= eu(ln a - ln b)
= euln a - uln b
= euln ae-uln b
= euln a/euln b
= au/bu
6. (d/dx)(au)
= (d/dx)euln a
= euln a(d/dx)(uln a)
= euln aln a(du/dx)
= (auln a)(du/dx)
7. Since F'(x) = f(x), working backwards:
(d/dx)(1/ln a)au + C
= (1/ln a)(auln a)(du/dx)
= au(du/dx)
so, ∫audu = (1/ln a)au + 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.