Inequality Problem from Spivak's Calculus: Chapter 1, Problem 4, Subproblem XI

AI Thread Summary
The discussion revolves around solving the inequality 2x < 8, with participants clarifying that it can be rewritten as 2^x < 2^3. The key point is recognizing that since the function 2^x is monotonically increasing, the inequality implies x < 3. There is some confusion about whether the solution should focus on integer values, but the consensus is that the solution set includes all real numbers less than 3. Participants appreciate the need for a clear mathematical context to support the proof.
Why?
Messages
6
Reaction score
0

Homework Statement


Find all numbers x for which:

2x<8

Homework Equations


The Attempt at a Solution



I really haven't been able to figure this one out.
 
Physics news on Phys.org
Do you have any calculations to show?

I assume it means integer values? If so surely isn't it just:

x=-\infty, ..., 0, \pm 1, \pm 2, \pm 3

Unless I'm missing the point somewhere? It seems a bit simple though.

:smile:
 
Note that 8=2^{3}, 2^{x} in monotonically increasing. So the question is, what values of x satisfy
<br /> 2^{x}&lt;2^{3}<br />
can you say what values satisfy this equation?
 
Stated in a different way, \log_2(t) is an increasing function. Inequalities remain true if you apply an increasing function.
 
Sorry I wasn't very clear.

Just thinking it through I know that 23 is 8, so x<3.

However my difficulty was in proving it, using the mathematical context that Spivak uses.

hunt_mat's example makes a lot of sense to me and fulfills that need to explain it more concretely.

Thanks for your help!
 
Back
Top