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X + 3^x < 4 ? Spivak got me on Chap. 1!

  1. Dec 5, 2007 #1
    Must be something missing from my repertoire-- Spivak got me in Chapter 1! :-)

    Trying to find all x that satisfy:

    [tex] x + 3^x < 4 [/tex]

    I've tried everything I can think of. Here are a few lines I've run down, to no avail:

    [tex] x + 3^x < 4 \Rightarrow e^{x+3^x} < e^4 \Rightarrow e^x \cdot e^{3^{x }}< e^4 \Rightarrow e^x \cdot e^{e^{x ln 3}} < e^4 [/tex] , and its more complicated.

    [tex] x + 3^x < 4 \Rightarrow 3^x < 4 - x \Rightarrow x \cdot ln 3 < ln (4-x) [/tex], and I'm unsure how to usefully proceed.

    What I thought to be most promising was:

    [tex] x + 3^x < 4 \Rightarrow x + e^{x ln 3} < 4 [/tex], unsure how to proceed.

    Does [itex] x + e^{ax} = b[/itex] have a general solution for x? What am I missing? I'm beginning to wonder if I should make some general arguments based around all [itex]e^x[/itex] being positive, or some such thing, after some manipulation.
     
  2. jcsd
  3. Dec 5, 2007 #2

    Office_Shredder

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    First notice if x decreases, x + 3x decreases. Then you just need to find where you get equality, and it's everything less than that point
     
  4. Dec 5, 2007 #3
    Yeah, I see qualitatively that that is the case. The problem is, I don't see how to solve for the exact value-- [itex] x + 3^x = 4[/itex] is no easier for me.

    I'm certain I could come up with the solution numerically in no time, but then, I'd still have this knowledge gap :P

    Thanks!
     
  5. Dec 5, 2007 #4

    Avodyne

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    Try guessing. x=0? x=2? Hmm ...
     
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