How can I minimize expression involving logarithms?

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)}}   *)

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} *)

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}}