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Posers and Puzzles

Posers and Puzzles

  1. 11 Jun '09 10:28
    Does the equation (2x)^lg2 = (3x)^lg3, x>0, has any rational solutions?
    If so, show me those in the form of p/q, and how you you came up with the answer.
    (Mathematica, and matlab, calculators, and other electronic devices shows only that you know how to press buttons, not mathematics.)
  2. Standard member Palynka
    Upward Spiral
    11 Jun '09 13:53
    Not a particularly elegant solution, but...

    Take logs and get:

    ln(2x)*ln(2)=ln(3x)*ln(3) <=>

    ln(2x)/ln(3) = ln(3x)/ln(2) <=>

    We can use the fact that ln(x) = - ln(1/x) and we see immediately that setting x=1/6 will give us -1 = -1 and is therefore a rational solution.

    In fact, let {a,b} be two non-zero constants, then ln(a*x)*ln(a)=ln(b*x)*ln(b) will always admit a solution at x = 1/(a*b). This is easily checked by plugging in the solution.

    Moreover, this solution is unique (if a =/= b) because of the strictly increasing and concavity properties of the log function.

    To see this note that ln(a.x)= ln(a)+ln(x) and that ln(a)*ln(x) and ln(b)*ln(x) never intercept, so once the above expressions intercept, they will never do so again.
  3. Standard member PBE6
    Bananarama
    11 Jun '09 14:24
    I was just going to add that you can solve for x explicitly:

    (2x)^ln(2) = (3x)^ln(3)

    Taking ln of both sides:

    ln(2)*(ln(2)+ln(x)) = ln(3)*(ln(3)+ln(x))

    Grouping ln(x) on the left:

    ln(x)*(ln(3)-ln(2)) = (ln(2))^2 - (ln(3))^2

    ln(x) = ((ln(2))^2 - (ln(3))^2) / ln(3/2) = K

    x = e^K

    Plugging in the numbers, we get x = 1/6 as Palynka noted above.
  4. 11 Jun '09 15:49
    Originally posted by PBE6

    ln(x)*(ln(3)-ln(2)) = (ln(2))^2 - (ln(3))^2
    You can get there a bit more elegantly from this point.

    ln(2)^2 - ln(3)^2 = [ln(2) - ln(3)][ln(2) + ln(3)]

    => ln(x) = -[ln(2) + ln(3)] = -ln(6) = ln(1/6)
  5. Subscriber joe shmo
    Strange Egg
    11 Jun '09 16:40 / 1 edit
    I'm curious as to how everyone magically changed from log base 10 to the natural log base?...or am I missing something?

    as far as i can see

    lg(x) =/= ln(x)
  6. Standard member Palynka
    Upward Spiral
    11 Jun '09 16:47
    Originally posted by joe shmo
    I'm curious as to how everyone neglected to change from log base 10 to the natural log base?...or am I missing something?

    as far as i can see

    lg(x) =/= ln(x)
    The result is the same, regardless of the base. If he meant log10, then just replace accordingly.
  7. Subscriber joe shmo
    Strange Egg
    11 Jun '09 16:59
    Originally posted by Palynka
    The result is the same, regardless of the base. If he meant log10, then just replace accordingly.
    ok,...so I was missing something.

    thanks
  8. Standard member Palynka
    Upward Spiral
    11 Jun '09 17:36 / 1 edit
    Originally posted by joe shmo
    ok,...so I was missing something.

    thanks
    You had a point, though. I did misread the initial problem. I'm not used to lg notation as I usually use log or ln for natural logarithm and log10 for base 10 (and logb for base b...). It just doesn't change as the actual base for the logarithm isn't used, just general logarithm properties.
  9. 11 Jun '09 18:10
    I see that you've solved the problem. Well done! The answer is of course 1/6, and using the logaritm laws it's quite easy.

    I didn't invent the problem, I used the same formulation as the source: lg. And as staed, the base doesn't matter, it gives the same result. I admit that I didn't think of that.