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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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    What conditions are set on *m* and *n*?2010-08-04

1 Answers 1

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There are many unnecessary variables. Let

$\alpha = \frac{an}{mb^2}.\qquad(1)$

Then the inequality becomes

$ 0 < 1 - \alpha e^{-r (T - t)} < 1 $

The first obvious step is perform “1 −” on every parts,

1 > \alpha e^{-r (T-t)} > 0

Since the exponential function's range is positive and < 1 (since r > 0 and T > t > 0), we can ensure α is positive.

If 0 < α ≤ 1, then every t will satisfy the inequality (the first solution).

So assume α > 1. Now it's pretty obvious on how to solve t in terms of r, T and α. Substitute (1) again to get back a, n, m and b.

Things you may consider:

  • ex is strictly increasing.
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    @Vaolter: Sorry that's not because of the condition *t* < *T*, but *t* > 0. So you get 0 < t < T - \frac1r \ln\alpha. Ignore the *t* and solve for *α*. (And I don't think $an = mb^2 e^{rT} is possible.)2010-08-05