Find all numbers x for wich:
[tex]x+3^x<4[/tex] Relevant equations
(PI) (Associative law for addition)
(P2) (Existence of an additive identity)
(P3) (Existence of additive inverses)
(P4) (Commutative law for addition)
(P5) (Associative law for multiplication)
(P6) (Existence of a multiplicative identity)
(P7) (Existence of multiplicative inverses)
(P8) (Commutative law for multiplication)
(P9) (Distributive law)
(P10) (Trichotomy law)
(P11) (Closure under addition)
(P12) (Closure under multiplication) THEOREM l For all numbers a and b, we have
[tex]\mid{a+b}\mid\leq\mid{a}\mid+\mid{b}\mid[/tex] I tried everything I'd already done with all the other problems but I just can't figure it out. Note that what I'm afteris the prove, I know what the result is. if you want look at it:

Solving x + 3^{x} = 4 for the exact solution is not something that is taught in precalculus courses, or even most calculus courses. About the best you can do is to get an approximate solution to x + 3^{x} = 4 (either graphically or by some estimation technique), and then use that to determine the interval for which x + 3^{x} < 4.

The best you can do here with elementary techniques is to sketch the curve y = x + 3^x, prove it's monotone increasing throughout, then hazard an intelligent guess that when y = 4, x = 1, which is trivially proven by substitution.

Yes, the problem is in Chapter 1 (Basic Properties of Numbers), No. 4 (xii).

Everything else in No. 4 is exactly solvable, except this one. There's nothing on this in the "Answers to selected problems" section either. I'm guessing this problem was included only because it has an "obvious" solution, which can be easily proven to be unique with elementary techniques like curve-sketching, which even a beginning student of Calculus is expected to be already familiar with.