One could recast this in Mathematica as below.

f[1] = 1;
f[n_] := f[n - 1]*x
Solve[f[10] == .03 && x > 0, x]

(* Out[234]= {{x -> 0.677315868387}} *)

Or just take (.3)^(1/9).

Hello, Kind of a related question. But what if I am trying to Repeat Calculations/Iterations without a specific function?

I am trying to repeat an initial condition many times such that I get a desired final output. ONE final output depends on many subsequent calculations from ONE initial condition. But I want to get a specific final output. I have many given parameters

Clear["Global`*"]

tid = 50; taid = 20; tcid = 30; rho = 0.975; pe1 = 4.37; pe2 = 4.4509; qedom = 1021167; 

qeim = 5700; qcid = 529.59; el = 0.655; did = 4.22; dtot = 430932.474; iad = 445002; 

cad = 688723.8554;

qetot = qedom + qeim;

uel = 1 - el;

sigid = 0.20568; rid = 0.02958;

upid = N[Exp[Sqrt[1]*sigid]];

dwid = N[1/upid];

rf = N[Exp[rid*1]];

pid = N[(rf - dwid)/(upid - dwid)]; 

qid = N[1 - pid];

prf2 = 0.001*pe2*qetot-0.01*cad; 

npva = ((rho - rho^(taid + 1))/(1 - rho))*prf2 - 0.01*iad;  

I would encourage you NOT TO CHANGE the values of the parameters(you are free to change the names if you wish); these are from real data and heavily influence the results. The following parameter "cprd" is the INITIAL CONDITION (given the other parameters) that I want to change many times

exch=63.56; cprd = 200; 

pin = exch*cprd;(*63.56 is a FIXED constant and "pin" is the product exch*cprd*)

prf1[p_] = 0.001*pe1*qedom*el - 0.001*p*qcid - 0.001*pe1*qeim;

As you see, ALL THE FOLLOWING calculations depends on ONE chosen value of "pin" and if I change "cprd" all the results would change.

stsp = Table[pin*(upid^i), {i, -tid, tid}];

Dimensions[stsp] (*checking this gives a column vector*)

{101}

lensp = Length[stsp]

101

prf1[stsp];

per[i_] := prf1[stsp][[i]];

up[n1_, sig_, T_] := N[Exp[Sqrt[T/n1] sig]];

down[n1_, sig_, T_] := N[1/up[n1, sig, T]];

int[n1_, Rf_, T_] := N[Exp[Rf*(T/n1)]];

P[up_, down_, al_] := N[(al - down)/(up - down)];

Q[up_, down_, al_] := N[1 - P[up, down, al]];

mean[l_List] := Apply[Plus, l]/Length[l];

AmericanOption[p0_, n1_, sig_, T_, Rf_, exercise_Function] := 
  Module[{u = up[n1, sig, T], d = down[n1, sig, T], 
    al = int[n1, Rf, T], p, q, OpRecurse, res}, p = P[u, d, al]; 
   q = Q[u, d, al]; 
   OpRecurse[node_, level_] := 
    OpRecurse[node, level] = 
     If[level == n1, exercise[p0*d^node u^(level - node)], 
      rho*{p, q}.{OpRecurse[node, level + 1], 
          OpRecurse[node + 1, level + 1]} + 
       exercise[p0*d^node u^(level - node)]]; res = OpRecurse[0, 0]; 
   Clear[OpRecurse]; res]; 

AmericanPut[p0_, n1_, sig_, T_, Rf_] := AmericanOption[p0, n1, sig, T, 
  Rf, # &] 

AmericanPut[per[1], tcid, sigid, tcid, rid]  (*just checking*)
95832.1

tabnpvc =Table[AmericanPut[per[i], tcid, sigid, tcid, rid], {i, 1, lensp}];

e1 = 0;

newp = Table[e1, {i, lensp}, {j, lensp}];

Dimensions[newp]

{101, 101}

Table[newp[[i, Min[i + 1, lensp]]] = pid, {i, lensp}]; 


Table[newp[[i, Max[i - 1, 1]]] = qid, {i, lensp}];

    valtern = Max[#, npva] & /@ tabnpvc;

The matrix "val" is the MOST IMPORTANT. I want to REPEAT THE ITERATIONS (by changing "cprd") such that difference between the 1st and the 50th column val[[All, 1]]-val[[All, 50]] is first positive and then zero (vector contains positive numbers followed by zeros).

val = Table[e1, {i, lensp}, {j, tid + 1}];

val[[All,tid + 1]] = tabnpvc; 

Do[Table[val[[All, i]] = Max[#, npva] & /@ (prf1[stsp] + rho*newp.val[[All, i + 1]]), {i, 
    1, tid}], {tid}];

val // MatrixForm; 

Dimensions[val];

diff = val[[All, 1]] - val[[All, 50]]; (*this vector should ideally contain positive values FOLLOWED BY zeros but NOT negative*)

pos = Position[diff, _?(# < 0 &)] (*this is not desired*)
{{36}, {37}, {38}, {39}, {40}, {41}, {42}, {43}, {44}, {45}, {46},{47}}

Finally, just checking with figures

valplot[i_, j_] := {stsp[[i]], val[[All, j]][[i]]}; 

val1 = ListLinePlot[Table[valplot[i, 1], {i, 1, lensp}], 
  PlotRange -> All]; 

val2 = ListLinePlot[Table[valplot[i, 50], {i, 1, lensp}], 
  PlotRange -> All]; 

Show[val1, val2]; (*this should be such that the curves don't ever cross but converge to the value of "npva"*)

So MY DESIRED FINAL OUTPUT is that the vector "diff" should contain BOTH positive numbers and zeros. It should first be positive and then zeros (maybe "pos" should equal the null vector {}). As you see, you can keep changing the value of "cprd" until you get the desired result

I have no idea on how to do incorporate all this in a "While" or "Do" function. Please help. Thank You.