Results 1 to 4 of 4

Math Help - Exponent Problem

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Exponent Problem

    My problem is this:
    Solve for t in terms of a, b, and c: a^t = bc^t

    So far I've gotten this:
    t loga = t logbc
    t loga = t(logb + logc)


    I don't know if that's right, and if that is right, where to go from there. I'm also not even sure what the problem is asking me to do! Do I need three solutions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    Solve for t in terms of a, b, and c: a^t = bc^t

    a^t = b(c^t), dividing both sides by c^t, you have now

    a^t / c^t = b, using the law of exponents, (x/y)^t = x^t/y^t, then

    (a/c)^t = b

    take logs of both sides,

    log (a/c)^t = log b, [use properties of log]

    (t) log (a/c) = log b

    (t) (log a - log c) = log b, cross-multiplying, you have

    t = (log b)/(log a - log c).


    lysserloo, i think this is the answer you are looking for....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    62
    AH! I never considered sectioning off c^t from b! I was treating it as (bc)^t, which it's not.

    Thank you SO much! I completely understand this now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    Yes, bc^t = b(c^t) not bc^t = (bc)^t.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponent Problem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: November 22nd 2011, 05:10 AM
  2. Replies: 3
    Last Post: April 27th 2011, 05:53 PM
  3. Replies: 1
    Last Post: August 19th 2008, 09:41 AM
  4. exponent problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 5th 2008, 07:42 PM
  5. Replies: 6
    Last Post: March 30th 2008, 11:39 AM

Search Tags


/mathhelpforum @mathhelpforum