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anonymous

  • 5 years ago

solve for x 7(1/2)^x=2x

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  1. amistre64
    • 5 years ago
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    gotta do logs and exponents :)

  2. amistre64
    • 5 years ago
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    take the log of each side: log[(7)(1/2)^x] = log(2x) ; now seperate the logs into addition. log(7) + log((1/2)^x) = log(2) + log(x)

  3. amistre64
    • 5 years ago
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    remember the log(AB) = log(A) + log(B)

  4. amistre64
    • 5 years ago
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    also log(N^x) = x log(N)

  5. anonymous
    • 5 years ago
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    ok thanks i think i can get it now

  6. anonymous
    • 5 years ago
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    Erm, unless I am missing something, that cannot be solved exactly; only through iterations.

  7. anonymous
    • 5 years ago
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    still cant solve it

  8. amistre64
    • 5 years ago
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    log(7) + x log(1/2) = log(2) + log(x) log(7) + x log(1) - x log(2) = log(2) + log(x) [x log(1) - x log(2)] -log(x) = log(2) - log(7) x log(1) - x log(2) ------- ------- = log(2/7) log(x) log(x) log[(1^x)/x] - log[(2^x)/x] = log(2/7) log [(1^x)/(2^x)] = = log(2/7) (1^x)/(2^x) = 2/7 7(1^x) = 2(2^x) this is my stuff so far :) might be helpful

  9. anonymous
    • 5 years ago
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    The answer is about 1.362, but it's unlikely you'll be able to solve it much better.

  10. amistre64
    • 5 years ago
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    maybe change of base? thats what I was thinking....

  11. anonymous
    • 5 years ago
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    It cannot be solved. You made a mistake.

  12. amistre64
    • 5 years ago
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    i probably did :) can you point it out?

  13. anonymous
    • 5 years ago
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    [x log(1) - x log(2)] -log(x) = log(2) - log(7) x log(1) - x log(2) ------- ------- = log(2/7) log(x) log(x) Wanna explain what you did here? I may be missing it, but rest assured there is a mistake somewhere, as what you got to can be solved (exactly), but what you started with cannot.

  14. anonymous
    • 5 years ago
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    Either way, the next line also doesn't follow.

  15. amistre64
    • 5 years ago
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    i was thinking along the lines of log(a) - log(b) = log(a/b) log(1^x) - log(2^x) ---------------- split the fraction? log(x)

  16. amistre64
    • 5 years ago
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    ooohh, I see it lol

  17. amistre64
    • 5 years ago
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    log[(1^x/2^x)/x] would that be right?

  18. anonymous
    • 5 years ago
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    newton has the right answer but im just not seeing how to solve for it

  19. amistre64
    • 5 years ago
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    another way of looking at it is: 1^x 2x ---- = --- 2^x 7

  20. amistre64
    • 5 years ago
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    yeah, its the playing around with the numbers and rules that I like to practice :)

  21. anonymous
    • 5 years ago
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    (1/2)^x = 2x/7 is right, but you can't go beyond that (in elementary functions for an exact answer; only iterations will get you a near-ish result)

  22. anonymous
    • 5 years ago
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    Oh, wait, of course that is right, it's almost exactly as given. Doh. But yeah, no solution, bad question

  23. anonymous
    • 5 years ago
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    im just lost on isolating the x. so say if you had 3^X=2X how would i isolate x

  24. amistre64
    • 5 years ago
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    lol :)

  25. amistre64
    • 5 years ago
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    take log3 of each side maybe as a start?

  26. amistre64
    • 5 years ago
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    x = log3(2x) x = log3(2) + log3(x) stuff like that

  27. anonymous
    • 5 years ago
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    You cannot do it. See ' Lambert W Function ' .

  28. amistre64
    • 5 years ago
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    or find a convient "base" and see if you can manipulate it...... its always fun to try the impossible :)

  29. anonymous
    • 5 years ago
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    Depending on the level, you are almost certainly meant to iterate this to get nearer and nearer to the root.

  30. amistre64
    • 5 years ago
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    3^3^X = 3^2X 3^3X = 3^2X .... is that allowed?

  31. anonymous
    • 5 years ago
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    Nope (and for the record, 3^x = 2x has no real solutions at all, regardless of method). 3^(3^2) = 3^9 3^(3*2) = 3^6

  32. anonymous
    • 5 years ago
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    (a^b)^c = a^(bc) =/= a^(b^c), in general

  33. anonymous
    • 5 years ago
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    Wow. So you can't use logarithms for this solution effectively? No solution.. agh. >.<

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