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Hard Algebra 2 Problem?

  1. Jan 21, 2012 #1
    x = 3^6 * 2^12
    y = 3^8 * 2^8
    x^x * y^y = z^z for some integer z

    Find z.

    I took the log of both sides, but don't know what to do next (I'm not even sure taking log of both sides will produce anything).
     
  2. jcsd
  3. Jan 21, 2012 #2

    I like Serena

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    Welcome to PF, de.bug! :smile:

    What about simply substituting x and y in x^x * y^y and simplifying?
     
  4. Jan 22, 2012 #3

    D H

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    That is a going to get more than a bit unwieldy. x and y are already fairly big numbers. x^x and y^y are incredibly large numbers.


    Hint: You need to make some assumption about the nature of z. What do the nature of x and y suggest? With the right assumption, taking the log of both sides will lead to the solution.
     
  5. Jan 22, 2012 #4

    I like Serena

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    It won't be unwieldy.
    As long as you don't actually calculate anything, but stick to powers of 2 and 3, the result is obtained in 6 lines.
     
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