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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)? – Casebash Aug 4 '10 at 1:17
For the first part, we have to also use $e^{-r(T-t)} \le 1$ (assuming that T>t, which you haven't stated) – Casebash Aug 4 '10 at 1:24
Casebash, I mean $X$ OR ($Y$ AND $Z)$ and you are right $T>t$, sorry I did not state that. – Vaolter Aug 4 '10 at 13:11
Which are the variables? – Mariano Suárez-Alvarez Aug 4 '10 at 15:57
What conditions are set on m and n? – kennytm Aug 4 '10 at 17:36

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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partly answered, how about $an \leq mb^2e^{rT}$? I dont get it the $t$ is gone. – Vaolter Aug 5 '10 at 1:01
@Vaolter: How did you arrive at that? Also, do not expand α until you've reached the step t > (or <) something. – kennytm Aug 5 '10 at 4:28
@KennyTM thats the solution i want to get. – Vaolter Aug 5 '10 at 10:22
@Vaolter: It is to ensure the solution $t < T - \frac1r\ln \alpha$ is consistent with the condition $t < T$. – kennytm Aug 5 '10 at 10:43
@KennyTM will you tell me how to get $an≤mb^2e^{rT}$? – Vaolter Aug 5 '10 at 15:29

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