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Algerbraic logarithm method question

  1. Sep 20, 2009 #1
    First of all, greetings to the scientific community here at Physics Forums.

    The following set of equations is given:
    y^(x-3y) = x^2
    x^(x-3y) = y^8

    With the next assumption given: x-3y is unequal to (-4).

    My attempt was to isolate the y variable in both of the equations so that i will be able to devide the the equations and recieve a simple x^a = b equations. This did not help, because the powers of the x's included the y, with no option to get rid of it.

    I Would like your assistance with finding the shortest, most efficient, non-logarithmic way to solve this question, asap.
     
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2

    Hurkyl

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    Why haven't you taken the logarithm of the equations? Isn't that the method you said you were asking about in the title?
     
  4. Sep 20, 2009 #3
    You are right it is a possibility. I did not fully mention my request here in the topic.
    I am searching for an additional method to logarithm, by modifying the powers in the two equations.

    Just tackled with another question with great similarity:

    x^(2x+y) = y^4
    y^(2x+y) = x^16


    I noticed the identity of the powers in the left sides of the equations, but even after immediatly deviding the two equations, I get stuck.

    What would be the most efficient way to solve this with no log?
     
    Last edited: Sep 20, 2009
  5. Sep 20, 2009 #4
    It is the method of logarithm, you just don't write log ...

    [itex] x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}[/itex]. Therefore [itex] (2x+y)^2 = 64[/itex] (that's where we really did logarithms without mentioning the word), so assuming [itex] x>0, y>0[/itex] we have [itex] 2x+y=8[/itex]. Then [itex] x^8 = y^4[/itex] so [itex] x^2=y[/itex]. Substitute in the equation: [itex]2x+x^2=8 [/itex] then solve the quadratic equation. The only positive solution is [itex]x=2[/itex] so then [itex]y=4[/itex].

    I assumed x and y are positive so that irrational exponents would be defined. But to solve for integer solutions, you could allow negative x,y since integer powers still make sense.
     
  6. Sep 20, 2009 #5
    edgar, using which formula did you make this calculation:

    [itex]
    x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}
    [/itex]
     
  7. Sep 22, 2009 #6
    Start with [itex] x^{(2x+y)} = y^{4}[/itex], raise both sides to power [itex] 2x+y[/itex] , get [itex] x^{(2x+y)^2} = y^{4(2x+y)} [/itex].
    Start with [itex] y^{(2x+y)} = x^{16}[/itex], raise both sides to the power [itex] 4[/itex] , get [itex] y^{4(2x+y)} = x^{16\cdot 4}[/itex] .
    OK?
     
  8. Sep 23, 2009 #7
    yes, I now understand this technique. very useful and widespread in problems involving equations.

    *According to this, when solving the quadratic equation, two answers apply: x=2, and x=-4.
    why is x=-4 (y=16) not valid in this case?

    ** It is also important to mention that another solution in this form of questions:
    when the two bases are equal to 1, the powers can be placed with any suggested number.
    Therefore: when x=1, y=1.
     
    Last edited: Sep 23, 2009
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