1
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Function is:

f[x_] = 1/Sqrt[Log[a, x - 1]]

The base of the logarithm contains parameter a, which means that in the process of manual calculation, a needs to be classified and discussed, in two cases: a>1 and 0<a<1. The result calculated using the following code is incorrect.

In[27]:= FunctionDomain[f[x], x]

Out[27]= 
x > 1 && Log[a] != 0 && Log[-1 + x] != 0 && Log[a] Log[-1 + x] > 0

the correct answer is:

if a>1,x>2
if 1>a>0,1<x<2
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1
  • $\begingroup$ And your question is...? You can report the behavior to Wolfram Support if you'd like, but I am not sure what answer you seek in this forum. $\endgroup$
    – MarcoB
    Jun 5, 2023 at 13:03

1 Answer 1

6
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The "correct" answer follows with Reduce

FunctionDomain[f[x], x] // Reduce[#, x] &
(*(Log[a] < 0 && 1 < x < 2) || (Log[a] > 0 && x > 2)*) 
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3
  • 1
    $\begingroup$ Make the argument of Reduce a instead of x to get the result the OP wants. $\endgroup$
    – Bill Watts
    Jun 5, 2023 at 22:06
  • $\begingroup$ In[5]:= FunctionDomain[{f[x]}, x] // Reduce[#, a] & Out[5]= (1 < x < 2 && 0 < a < 1) || (x > 2 && a > 1) $\endgroup$
    – csn899
    Jun 5, 2023 at 23:01
  • 1
    $\begingroup$ @BillWatts OP 's correct answer contains parameter a at the first place $\endgroup$ Jun 6, 2023 at 7:08

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