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Can someone explain why the term A in case 2 does not get canceled out when case 1 does? I use "9.0 for Microsoft Windows (64-bit) (January 25, 2013)"

case1 = A*(E^(-λ) + α) == A;
Simplify[case1, A > 0]

gives 1 + E^λ α == E^λ

case2 = A*(E^(-2*λ) + α) == A;
Simplify[case2, A > 0]

gives A (E^(-2 λ) + α) == A

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I get E^(-2*λ) + α == 1 for case2 in version 7. Perhaps try FullSimplify? – Mr.Wizard Jan 25 '14 at 0:06
@Mr.Wizard And also in version 8. But not in version 9! Moreover, case1 is pretty weird too because E^-\[Lambda] + \[Alpha] == 1 (what v8 returns) is clearly simpler than 1 + E^λ α == E^λ (what v9 returns). A v9 bug perhaps? user11946: you can consider reporting this to support at since it seems that both results you mention are worse than what v8 (or v7) gives. – Szabolcs Jan 25 '14 at 0:12
I tried. FullSimplify does not do it in version – user11946 Jan 25 '14 at 0:13
@user11946 Why don't you write to support about this problem and report back here with what they told you? – Szabolcs Jan 25 '14 at 0:14
Yes, I will report it to Wolfram. – user11946 Jan 25 '14 at 0:19

Below is the email response from Wolfram support.

Thank you for your email. It is not clear to me why Mathematica's behavior has changed between version 8 and 9 in this example with Simplify. As such, I have filed a report with our developers so that they can be aware of this behavior and make any changes as needed for future versions of Mathematica. In the meantime, I will also be sure to pass on any useful information that I get from our developers. Please let me know if you have any further questions.

Karl Isensee

Technical Support Wolfram Research, Inc.

share|improve this answer
See this answer to get an insight how to proceed. – Artes Jan 27 '14 at 23:27

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