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.

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