1
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How do I get

Solve[(1/9)^(r/2) == 1/n , r] 

to give me

 Log[n]/Log[3]

and not

-Log[1/n]/Log[3]?

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  • $\begingroup$ Thanks. Solve may have just been the wrong command. $\endgroup$ – lip1 Dec 11 '12 at 15:18
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    $\begingroup$ @b.gatessucks Perhaps you should consider posting working answers, however short they may be, as answers and not as comments. The list of seemingly unanswered questions is growing and growing. I won't hold it against you ;-) $\endgroup$ – Sjoerd C. de Vries Dec 11 '12 at 15:35
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    $\begingroup$ You can get the result with Solve, you can do e.g. this Simplify[ Solve[ (1/9)^(r/2) == 1/n, r, Reals], n > 0]. $\endgroup$ – Artes Dec 11 '12 at 15:37
  • $\begingroup$ @SjoerdC.deVries the alternative is someone collects all these answers in comments and produce 200 answers ;-) $\endgroup$ – chris Dec 11 '12 at 15:44
  • $\begingroup$ Or I can answer my own question with the comments :) $\endgroup$ – lip1 Dec 11 '12 at 15:45
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This does it :

Reduce[(1/9)^(r/2) == 1/n && n > 0, r, Reals]

(* n > 0 && r == Log[n]/Log[3] *)
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