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Solving simultaneous equations

  1. Oct 13, 2014 #1
    I am trying to solve the simultaneous equation 2^x+y = 6^y and 3^x = 6(2^y) i have solved as follows
    3^x . 2^x+y = 3^x . 6^y
    3^x. 2^x+y = 6(2^y). 2^x+y on subtraction
    i get
    3^x.6^y - 6(2^y).2^x+y = 0 then
    3^x. 6^y = 6(2^y). 2^x. 2^y on solving i get
    x log (3/2)= log 6+y log(3/2) now i do i get x and y from this step. the answers are x=2.71 and y = 1.71 regards
  2. jcsd
  3. Oct 13, 2014 #2

    Char. Limit

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    Okay, let's see. So just to make sure I have the problem clear... your two equations are [itex]2^{x+y} = 6^{y}[/itex] and [itex]3^{x} = 6 \cdot 2^{y}[/itex]. Fun looking problem, really. Well, let's see... if you multiply both sides of the second equation by 2^x, you get this:

    [tex]6^{x} = 6 \cdot 2^{x+y}[/tex]

    Then if we substitute the second equation in for the first, we get this:

    [tex]6^{x} = 6 \cdot 6^{y}[/tex]

    Now this looks nice and simple. The right side simplifies to 6^(y+1), and so we have 6^(x) = 6^(y+1). Now, there's a certain property, I can't remember what it is, but basically it says if a^x = a^y, then x=y. So using that property, we have x=y+1. Good! We found a simple linear relation between x and y. Now we just need to figure out what satisfies this. To solve that, we substitute y+1 wherever we see an x in the first equation to find the solution for x, which we then need to subtract 1 from that solution to get y. That should be the whole solution set.
  4. Oct 13, 2014 #3
    Thanks i now get it
    2^x+y =6^y since x=y+1,
    2^y+1+y = 6^y
    2^2y+1 = 6^y
    4^y+2 = 6^y
    log 2 = y log (6/4)
    y= log 2/ log 1.5
    y= 1.709 hence x = 1.709+1 = 2.709
  5. Oct 13, 2014 #4


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    The property is that the function is "one-to-one": if f is a "one-to-one" function and f(x)= f(y) then x= y. Or, equivalently, that the exponential function, [itex]a^x[/itex], has the inverse function, [itex]log_a(x)[/itex]. If [itex]a^x= a^y[/itex] then [itex]ln(a^x)= ln(a^y)[/itex] so that x= y.
  6. Oct 13, 2014 #5

    Char. Limit

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    Ah, thank you, that's the one! Of course, the exponential function is only one-to-one if we assume x and y are both real, but we can safely assume that here.
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