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

1. Dec 5, 2007

### dotman

Must be something missing from my repertoire-- Spivak got me in Chapter 1! :-)

Trying to find all x that satisfy:

$$x + 3^x < 4$$

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

$$x + 3^x < 4 \Rightarrow e^{x+3^x} < e^4 \Rightarrow e^x \cdot e^{3^{x }}< e^4 \Rightarrow e^x \cdot e^{e^{x ln 3}} < e^4$$ , and its more complicated.

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

What I thought to be most promising was:

$$x + 3^x < 4 \Rightarrow x + e^{x ln 3} < 4$$, unsure how to proceed.

Does $x + e^{ax} = b$ 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 $e^x$ being positive, or some such thing, after some manipulation.

2. Dec 5, 2007

### Office_Shredder

Staff Emeritus
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

3. Dec 5, 2007

### dotman

Yeah, I see qualitatively that that is the case. The problem is, I don't see how to solve for the exact value-- $x + 3^x = 4$ 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!

4. Dec 5, 2007

### Avodyne

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