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Math Help - Solving equation with logs and variables

  1. #1
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    Solving equation with logs and variables

    Problem: F(x) = 4^(x) for all real values of x. If p>1 and q>1, then f^(-1)(p)f^(-1)(q)=?

    The notation above for f^(-1) means inverse function.

    The answer is log(base 4) p x log(base 4)q

    I first attempted to solve for the inverse function by switching y and x and then solving for y. This did not get me log (base4) x, which the book says is the inverse of the exponential function.

    Last edited by benny92000; November 27th 2011 at 05:01 PM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Solving equation with logs and variables

    Can you show how you came to your answer?
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    Re: Solving equation with logs and variables

    I didn't come to the right answer, but I have y=4^(x) I switched x and y to solve for the inverse. x=4^(y) I took the log of both sides logx= ylog4. Then I got logx/log4. I then plugged in p and q and fiddled with it from there to no avail.
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    Re: Solving equation with logs and variables

    I assume again ...

    f(x) = 4^x (you used a capital F)

    you should know that the inverse of an exponential function is a log function, but here is the derivation anyway ...

    y = 4^x

    x = 4^y

    \log_4(x) = \log_4(4^y)

    \log_4(x) = y

    so ...

    f^{-1}(x) = \log_4(x)

    f^{-1}(p) \cdot f^{-1}(q) = \log_4(p) \cdot \log_4(q)
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    Re: Solving equation with logs and variables

    Quote Originally Posted by benny92000 View Post
    I didn't come to the right answer, but I have y=4^(x) I switched x and y to solve for the inverse. x=4^(y) I took the log of both sides logx= ylog4. Then I got logx/log4. I then plugged in p and q and fiddled with it from there to no avail.
    Which logarithm did you use? a^x gives a different function for every different a, so its inverse log_a(x) is a different function for every a. If you had used log_4(x), Since 4= 4^1, log_4(4)= 1 so your "log x/log 4" is just log_4(x). For any other logarithm, it is true that log_a()x)/log_a(4)= log_4(x). That's a property of logarithms worth knowing:
    \frac{log_a(x)}{log_a(b)}= log_b(x)
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  6. #6
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    Re: Solving equation with logs and variables

    Change of base rule. I am aware of it. I see how that applies here now. Thanks!
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