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Hi, I'm in AP Calculus and I feel really stupid

  1. Sep 3, 2003 #1
    Hi, I'm in AP Calculus and I feel really stupid :(

    I feel really stupid because it's only the first week of school and I'm struggling with the review (I forgot a lot over the summer)

    I'm posed with these two problems (in addition to many other but I don't want to bore you all :)):

    (a)Solve for x: x^2 = 2^x
    (b)Solve for X: x^2 > 2^x

    My first plan of attack for (a) was to do this:

    2logx = xlog2

    that didn't seem to get me anywhere and now I seem stuck... please help me?

    Thanks,

    Daniel
     
  2. jcsd
  3. Sep 3, 2003 #2
    Also, does anyone know any forums similar to these (not physics though) which can help you with math homework? I posted here because I'm sure a lot of you are in high levels of math too.
     
  4. Sep 3, 2003 #3

    jcsd

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    Does re-arranging a) into xx = 22 help you?
     
  5. Sep 3, 2003 #4

    Hurkyl

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    The only real way to do (a) is by inspection. Make a few guesses as to the answer, you'll probably hit upon the two solutions.


    As for (b), are you reviewing calculus or previous courses? Are you allowed to solve problems like this by looking at the graph in a calculator?
     
  6. Sep 3, 2003 #5
    I tried graphing too, and that gave me 3 different intersections when I set y=x^2 and y=2^x. I still wouldn't know how to do (b) with graphing, and I'm sure there's a way to do (a) without graphing.
     
  7. Sep 3, 2003 #6

    jcsd

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    There is only one intersection for those two lines.
     
  8. Sep 3, 2003 #7
    I found an intersection at 2, -.7666, and one at 4. I used ti83+ and I'm in degree mode.
     
  9. Sep 3, 2003 #8
    We are reviewing previous courses. I'm allowed to use a calculator, but I forgot how to with a >.
     
  10. Sep 3, 2003 #9

    jcsd

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    oops sorry I've mad a mistake.
     
  11. Sep 3, 2003 #10

    jcsd

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    just to correct my first post, this is were I made the mistake, i should of gotten:

    x1/x = 21/2
     
  12. Sep 3, 2003 #11

    jcsd

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    The only way your going to solve this I think is by using your graphing calculator.
     
  13. Sep 3, 2003 #12
    how about part (b)?
     
  14. Sep 3, 2003 #13

    jcsd

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    b) should be 2 < x < 4 and x < -0.7666
     
  15. Sep 3, 2003 #14

    jcsd

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    basically again just looking at the graph.
     
  16. Sep 3, 2003 #15

    HallsofIvy

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    You "solve" inequalities with a graph by seeing which graph is above the other. You determine where each interval ends by "zooming" in on the intersections.

    You cannot solve equation like x^2= 2^x (which has x both as a "base" and as an exponent) algebraically (except by using the "Lambert W function" which is specifically defined as the inverse of the function f(x)= xe^x.)
     
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