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How to compare sizes/orders using a logarithmic equality?

Wen Dao
Posted 13 days ago

enter image description here

Reduce[Exists[{x, y, z}, {x > 0, y > 0, z > 0, 
     2 + Log[2, x] == 3 + Log[3, y] == 5 + Log[5, z], #}], 
   Reals] & /@ {x > y > z, x > z > y, y > x > z, y > z > x}

Since the solutions to logarithmic equations are not unique, it's impossible to determine a single exact solution for the equation. The method mentioned above cannot verify the correctness of the third option (Option C). Are there alternative approaches that can perfectly resolve this issue?

enter image description here
POSTED BY: Wen Dao
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Using FindInstance we see that all four relationships have exceptions. For condition A:

eqs = {x > 0, y > 0, z > 0, 
   2 + Log[2, x] == 3 + Log[3, y] == 5 + Log[5, z]};
{condsA, condsB, condsC, condsD} = {x > y > z, x > z > y, y > x > z, 
   y > z > x};
FindInstance[Append[eqs, Not[condsA]], {x, y, z}]
Append[eqs, Not[condsA]] /. % // FullSimplify

For condition B:

FindInstance[Append[eqs, Not[condsB]], {x, y, z}]
Append[eqs, Not[condsB]] /. % // FullSimplify

For condition C:

FindInstance[Append[eqs, Not[condsC]], {x, y, z}]
Append[eqs, Not[condsC]] /. % // FullSimplify

For condition D:

FindInstance[Append[eqs, Not[condsD]], {x, y, z}]
Append[eqs, Not[condsD]] /. % // FullSimplify
POSTED BY: Gianluca Gorni
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