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equation with two inequalities

find_the_fun

Active member
Feb 1, 2012
166
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
Jan 30, 2012
74
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)
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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