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.

Log in or register to write something here or to contact authors.