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This is simple example:

 v = {1.48622576, 1.53444276, 1.42299438}
 v/(v + 1)
 Out[1]:= {0.597784, 0.605436, 0.587288}

In the above case expression v/(v + 1) is in explicit form. I would like to find k for v = {1.48622576, 1.53444276, 1.42299438} wherein

 Log(0.5(1-k))/Log(0.5(1+k))=v (expression in implicit form)
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2
  • $\begingroup$ You have to look up how to properly set-up equations and correct your mistakes /syntax/. $\endgroup$
    – Sektor
    Apr 15, 2014 at 11:53
  • $\begingroup$ k /. Table[FindRoot[ Log[0.5 (1 - k)]/Log[0.5 (1 + k)] == v, {k, 1/2}], {v, {1.48622576, 1.53444276, 1.42299438}}] Out[103]= {0.136539357217, 0.14739964925, 0.121702474657} $\endgroup$ Apr 15, 2014 at 15:07

1 Answer 1

2
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As Daniel Lichtblau said in his comment the following works:

vs = {1.48622576, 1.53444276, 1.42299438};
eq[k_] := Log[0.5 (1 - k)]/Log[0.5 (1 + k)]
ks = k /. Table[FindRoot[eq[k] == v, {k, 1/2}], {v, vs}]
{0.136539, 0.1474, 0.121702}

Then you can check if the answer is correct:

vs == eq /@ ks
True
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