For what x does x + 3^x < 4

[zeion]

1. The problem statement, all variables and given/known data

For what x does x + 3^x < 4


2. Relevant equations



3. The attempt at a solution

The thing is = 4 when x is 1. So I want x < 1.
But is there a way to do this algebraically? Like with log or something?

[fzero]

Taking a log doesn't really make this problem much more transparent. If this were the Calculus forum, I'd suggest looking at Newton's method for finding a numerical approximation: http://en.wikipedia.org/wiki/Newton%27s_method

[Mark44]

You can say something about this inequality just by knowing something about the graphs of y = x and y = 3x. Both of these are strictly increasing functions, so their sum is also a strictly increasing function. Let's define f(x) = x + 3x.

Looking at asymptotic behavior, as x gets more and more negative, f(x) approaches the graph of y = x. IOW, for very negative x, f(x) \approx x. As x gets larger and larger, f(x) approaches y = 3x.

You already found out that f(1) = 4, so for any x > 1, then f(x) > 4. Similarly, if x < 1, f(x) < 4.