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Question: I want to solve $0<1−an/(mb^2)e^{−r(T−t)}<1$, where $r, a, b, T, t>0$.

The solution is that either $$an\leq mb^2$$ or $$mb^2\leq an\leq mb^2e^{rT}$$ and $$t< T − (\ln(an) − \ln(mb^2 ))/r$$.

My Attempt: My thoughts are that the first part $0<1−\frac{an}{mb^2}e^{−r(T−t)}$ gives me $an \leq mb^2$ because $e^{−r(T−t)}>0$, so I have the first part. The second part $1−\frac{an}{mb^2}e^{−r(T−t)}<1$ does not give me useful information since $\frac{an}{mb^2}e^{−r(T−t)}>0$ always.

How do I get the other half of the solution ( $mb^2\leq an\leq mb^2e^{rT}$ and $t< T − (\ln(an) − \ln(mb^2 ))/r$)?

I also realise that the problem I have to solve reduces to solving $xy<1$ where both $x,y>0$.


Merged from: tricky inequality

How do I go about solving $0<1−\frac{an}{mb^2}e^{−r(T−t)}<1$, where a,b,T>t>0? I have been stuck here for some time now.

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    Your second sentence effectively says X OR Y AND Z. Do you mean (X OR Y) AND Z or do you mean X OR (Y AND Z)?2010-08-04
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    For the first part, we have to also use $e^{-r(T-t)} \le 1$ (assuming that T>t, which you haven't stated)2010-08-04
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    Casebash, I mean $X$ OR ($Y$ AND $Z)$ and you are right $T>t$, sorry I did not state that.2010-08-04
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    Which are the variables?2010-08-04
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    I am looking for all the possibilities of $t$ in terms of $a,b,m, T$ or $a$ in terms of $b,m,T$2010-08-04
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    Can you explain that in the question?2010-08-04
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    And what have you tried? It should at least be obvious how to get it down to an inequality for $e^{-r(T-t)}$.2010-08-04
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    What conditions are set on *m* and *n*?2010-08-04

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