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Inverse of Exponential Equations

  1. Jun 12, 2007 #1
    Alright, I'm just going through my older tests in preparation for my upcoming exams and I have come across a question regarding exponents and inverses.

    1. The problem statement, all variables and given/known data

    The original equation is:

    [itex] h(x) = 3(4^{(x-1)})+2 [/itex]

    and I must find the inverse in order to proceed with the rest of the question.

    2. Relevant equations

    Now I know the basic stuff of Logs, Natural Logarithms and Exponents such as the basic rules:

    Change of Base

    and I can do the other questions on the test but it is this one question with the additional constants that is throwing my mathematical abilities out of the window:yuck:

    If someone could just give me some guidance on how to approach this problem I would be eternally grateful.

    - Murdoc
    Last edited by a moderator: Jun 12, 2007
  2. jcsd
  3. Jun 12, 2007 #2
    Also inside the brackets it is suppose to be 4^(x-1). I'm still new with the Latex coding
  4. Jun 12, 2007 #3


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    So you have some equation of the form [itex]y=3(4^{(x-1)})+2[/itex]. To find this inverse of this function, you need to make x the subject of the formula. So, firstly, subtract 2 and divide by 3 to give [tex] \frac{y-2}{3}=4^{(x-1)} [/tex]. Now, you can take the natural log of both sides, and use rules of logarithms to simplify. Alternatively, you could use logarithm with base four. Can you go on from here?
  5. Jun 12, 2007 #4
    I'm still confused as I've never done a question to this extent before, so it's still a relatively new concept to understand.

    If I take the Log base 4 route I end up with this:

    (x-1) = Log(4)((y-2)/3)
    X-1 = -3Log(4)(Y-2)

    Is this correct or am I missing something? and if that's correct, where would I go from here?

    Last edited: Jun 12, 2007
  6. Jun 12, 2007 #5
    log a/b = log a - log b
  7. Jun 13, 2007 #6


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    To express shramana's point more succintly; this line is correct:
    [tex]x-1 = \log_4(\frac{y-2}{3})[/tex]
    However, your next line is incorrect. You could simplify this using the logarithm of quotients rule then add 1 to both sides, or could just leave it as log((y-2)/3) and add 1 to both sides. It's up to you really.
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