- #1
dotman
- 127
- 0
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.
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.