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Is it possible to simplify equations like the following

  1. Dec 26, 2012 #1
    These are just some example equations:


    where the x is raised to the power. How can (if possible) I simplify these equations?
  2. jcsd
  3. Dec 26, 2012 #2
    I don't see any equations. An equation should have an "=" sign somewhere.
  4. Dec 27, 2012 #3


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    If you treat it as an expression (and not an equation like micromass pointed out), you might want to consider that for x > 0, y > 0,

    (SQRT(x))^(2a) - (SQRT(y))^(2b)

    = (SQRT(x)^a + SQRT(y)^b)*(SQRT(x)^a - SQRT(y)^b)
  5. Dec 27, 2012 #4


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    So generally, you're looking at expressions of the form [itex]a^x-b^x[/itex] for positive a and b.
    Sadly, no. That is the simplest form you can have it in.
  6. Dec 27, 2012 #5


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    You can factorise them, using the ordinary rules of numbers raised to powers, e.g. the first would be
    12x(5x - 3x) . Whether you call that a simplification and whether and when it is of any usefulness are other questions, but it shouldn't be a difficulty to see.
  7. Dec 27, 2012 #6
    Merci beaucoup à tout! I suspected that it wouldn't be able to reducible, but I just wanted to make sure.
  8. Dec 28, 2012 #7


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    [tex](6(10))^x- (6(6))^x= 6^x10^x- 6^x6^x= 6^x(10^x- 6^x)[/tex]

    [tex](6(5))^x- (5(5))^x= 5^x6^x- 5^x5^x= 5^x(6^x- 5^x)[/tex]

    However, neither [itex]10^x- 6^x[/itex] nor [itex]6^x- 5^x[/itex] can be further simplified.

    You mean "where the x is the power."

  9. Jan 3, 2013 #8
    If you wanted to solve an equation in this form (e.g. set it equal to something like a constant) you could solve it with the Lambert W function.
  10. Jan 3, 2013 #9

    D H

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    The first one can. The gcd of 10 and 6 is 2.
  11. Jan 3, 2013 #10

    D H

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    Previous posts have simplified these by factoring out the greatest common denominators of 60, 36 and of 30, 25.

    Another way is to take advantage of the fact that 1x=1:

    =& 36^x \left( (60/36)^x - (36/36)^x \right)
    = 36^x \left( (5/3)^x - 1\right) \\
    =& 25^x \left( (30/25)^x - (25/25)^x \right)
    = 25^x \left( (6/5)^x - 1\right)
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