How can I minimize expression involving logarithms?

Question

I am trying to find minimum of this expression:
$$log _aleft(frac{4}{9} (3b-1)right)+8 log_{frac{b}{a}}^2(a)-1,$$
where $ 0 < b < a < 1$. I tried
Clear[a, b]
    Minimize[Log[a, 4 (3 b - 1)/9] + 8 (Log[b/a, a])^2 - 1, 
     0 < b < a < 1, {a, b}]

But I don't get the result. How can I get the result?

Artes · Accepted Answer

We have a transcendental function of two variables and it is not straightforward to find global minima under given constraints because Minimize behind the scene uses symbolic equation solving functionality and sometimes it has to be supported by user's insight. However using both numerical and symbolic approach we can find an exact global minimum.

We define

f[a_,b_]:= Log[a, 4 (3 b - 1)/9] + 8 (Log[b/a, a])^2 - 1

To get an insight we play with

MinimalBy[ Table[ FindMinimum[{f[a, b], 0 < b < a < 1}, {b}], 
                  {a, 73/100, 95/100, 2/100}], First, 3]

{{7.00101, {b -> 0.659199}}, {7.02367, {b -> 0.702245}}, {7.04024, {b -> 0.619364}}}

By direct inspection we find out where we should look for the global minimum:

RegionPlot[{ f[a, b] < 7.01, f[a, b] < 7.001, f[a, b] < 7.0001}, 
            {a, 0.84, 0.9}, {b, 0.64, 0.7}, AxesLabel -> Automatic,
            WorkingPrecision -> 30, PlotPoints -> 60, MaxRecursion -> 5]

With quite a good numerical approximation we can find

FindMinimum[{f[a, b], 3/5 < b <= 4/5, b < a < 1}, {{a, 0.87}, {b, 2/3}}]

 {7., {a -> 0.87358, b -> 0.666667}}

similarily works NMinimize[{f[a, b], 3/5 < b <= 4/5 < a < 1}, {a, b}], while Minimize doesn't work this way, nevertheless restricting one variable we can find an exact result. It is obvious that both partial derivatives have to vanish in the extremum:

Solve[Derivative[0, 1][f][a, 2/3] == 0 && 1/2 < a < 1, a]

 {{a -> (2/3)^(1/3)}}

Minimize[{f[(2/3)^(1/3), b], 1/3 < b < 1}, b] // FullSimplify

{7, {b -> 2/3}}

Bob Hanlon · Answer

Clear["Global`*"]

f[a_, b_] := Log[a, 4 (3 b - 1)/9] + 8 (Log[b/a, a])^2 - 1

min = (FindMinimum[{f[a, b], 1/2 < b < a, 0 < a < 1}, {a, b}, 
  WorkingPrecision -> 20] // N) /. x_Real :> RootApproximant[x] // 
   ToRadicals

(* {7, {a -> (2/3)^(1/3), b -> 2/3}} *)

Verifying that the approximated results are exact

{min[[1]] == f[a, b], D[f[a, b], a] == 0, D[f[a, b], b] == 0} /. min[[2]]
  // FullSimplify

(* {True, True, True} *)

Answered by Bob Hanlon on August 25, 2021

Akku14 · Answer

You do not need numerical guess to find the minimum. Setting both partial derivatives to zero and eliminating variable a is enough.

f = Log[a, 4 (3 b - 1)/9] + 8 (Log[b/a, a])^2 - 1 // 
   PowerExpand[#, Assumptions -> {0 < a < 1, 1/3 < b < a}] &;

ee1 = (D[f, a] // Together // Numerator) // 
   PowerExpand[#, Assumptions -> {0 < a < 1, 1/3 < b < a}] &;

ee2 = (D[f, b] // Together // Numerator) // 
   PowerExpand[#, Assumptions -> {0 < a < 1, 1/3 < b < a}] &;

eli = Eliminate[{ee1 == 0, ee2 == 0}, Log[a]]

(*   b Log[b]^5 (6 Log[2] - 6 Log[3] - 3 Log[b] + 3 Log[-1 + 3 b]) == 
     Log[b]^5 (2 Log[2] - 2 Log[3] + Log[-1 + 3 b])   *)

Solve[eli && 1/3 < b < 1, b]

(*   {{b -> 2/3}}   *)

Solve[0 == (ee1 /. b -> 2/3) && 1/3 < a < 1, a]

(*   {{a -> (2/3)^(1/3)}}   *)

Answered by Akku14 on August 25, 2021

How can I minimize expression involving logarithms?

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