equation with two inequalities

find_the_fun

Active member
This is part of a larger problem and I'm trying to solve 2h+1≤n≤2h+1 for h. If the equation had two equals signs or one inequality I think I could do it but I'm not sure how to proceed with both. In other words, I don't know how to manipulate an expression of the form (expr1)≤(expr2)≤(expr3). Any help would be appreciated.

earboth

Active member
This is part of a larger problem and I'm trying to solve 2h+1≤n≤2h+1 for h. If the equation had two equals signs or one inequality I think I could do it but I'm not sure how to proceed with both. In other words, I don't know how to manipulate an expression of the form (expr1)≤(expr2)≤(expr3). Any help would be appreciated.
You can split a "chain of inequalities" into single inequalities connected by an $\wedge$-sign:

$2h+1≤n≤2^{h+1}~\implies~ 2h+1 \leq n ~\wedge ~ n\leq 2^{h+1}$

The final set of solutions is the interception of the two single sets of solutions (sorry if my English sounds a little bit cryptic)

• Jameson

CaptainBlack

Well-known member
This is part of a larger problem and I'm trying to solve 2h+1≤n≤2h+1 for h. If the equation had two equals signs or one inequality I think I could do it but I'm not sure how to proceed with both. In other words, I don't know how to manipulate an expression of the form (expr1)≤(expr2)≤(expr3). Any help would be appreciated.
Assuming both $$h$$ and $$n$$ are real:

Do each seperatly so $$2h+1\le n \Rightarrow h\le \frac{n-1}{2}$$, and $$n\le2^{h+1}\Rightarrow h \ge \log_2(n)-1$$

The simultaneous solutions are values in the overlap of the two parts of the real line to which the two conditions constrain $$h$$

CB

• earboth and Jameson