The function is simple enough, when you allow floating point numbers:

function getNums(n,f,tot){
let x=1;
 const arr=[...new Array(n)]
  .map(c=>x*=f);
 const mul=tot/arr.reduce((a,c)=>a+c);
 return arr.map(v=>mul*v)
}
 
console.log(getNums(3, 1.5, 100))

In order to get a unique result you need to specify the desired count n of numbers, the factor between the numbers f and the total tot the individual numbers must add up to.

And here is an adapted solution for integers:

function getNums(n,f,tot){
let x=1,res;
 const arr=[...new Array(n)]
  .map(c=>x*=f);
 const mul=tot/arr.reduce((a,c)=>a+c);
 
 res=arr.map(v=>Math.round(mul*v));
 res[res.length-1]-=res.reduce((a,c)=>a+c)-tot;
 return res

}
const ar=getNums(3, 1.5, 100);
console.log(ar,ar.reduce((a,c)=>a+c))

@Redu provided a formula for calculating the sum of the series 2**i for n values . By using the general annuity formula you will find that a suitable starting value x can be calculated through the following equation:

When I apply this formula I can save a few calculations in my script:

// small utility function
function sum(a,c){return a+c}

// create a series of n integers with 
// factor b between them, having a total sum of tot:
function numSeries(b,n,tot){
 let x=tot*(1-1/b)/(b**n-1)

 const ar=[...new Array(n)]
  .map(_=>Math.round(x*=b) );
  // adjust the last number for rounding errors:
 ar[n-1]=tot-ar.slice(0,-1).reduce(sum);
 return ar;
}

// test it all:
const ar=numSeries(1.75,7,200);
console.log(ar,ar.reduce(sum))

This is a Math problem. Once you solve it, implementing this with JS or any other language is just calculation.

So you want K numbers summing to T. However they will increase exponentially. Not so clear but from your given example lets assume our exponential series to be

T = x + 2x + 4x + 8x + ... + 2^(n-1) x

We can apply summing series trick here like

2T = 2x + 4x + 8x + ... + 2^n x

Now;

   2T = 2x + 4x + 8x + ... + 2^x
    T =  x + 2x + 4x + 8x + ... + 2^(n-1) x
(-)__________________________________
    T = 2^n x - x = (2^n - 1) x

So;

        T
x = _________
     2^n - 1

The JS part.

function part(n,target){
  var x = target / (2**n-1);
  return Array.from({length:n}, (_,i) => x*2**i);
}

var result = part(6,100),
    sum    = result.reduce((p,c) => p+c);

console.log(result,sum);

This should do it. The exponential function is: A^numParts = desiredTotal. We first solve for A then push the values calculated values into the array. This works because the integral of an exponential function is the exponential function.

function returnParts (desiredTotal, numParts) {
  var a = Math.pow(desiredTotal, 1/numParts)
  var parts = []
  for (var j = 0; j < numParts; j++) {
    var part = Math.pow(a, j + 1) - Math.pow(a, j)
    parts.push(part)
  }
  return parts
}

For example:

var results = returnParts(1000, 3)
console.log(results)
//
//returns [9.999999999999998, 89.99999999999997, 909.9999999999994]
// which is equal to ~1000
//