Question 1209869
Let's break down this problem step-by-step.

**1. Understand the Domain**

First, we need to find the domain of the function inside the logarithm. We need:

* `-20x + 16√x - x > 0`
* `-21x + 16√x > 0`
* `16√x > 21x`

Since `√x` and `x` are involved, we know that `x ≥ 0`.

Let `y = √x`, so `x = y²`. Then the inequality becomes:

* `16y > 21y²`
* `16y - 21y² > 0`
* `y(16 - 21y) > 0`

This inequality holds when `0 < y < 16/21`. Since `y = √x`, we have:

* `0 < √x < 16/21`
* `0 < x < (16/21)²`
* `0 < x < 256/441`

So the domain is `0 < x < 256/441`.

**2. Maximize the Inside of the Logarithm**

Since the logarithm is an increasing function, maximizing `f(x)` is equivalent to maximizing the expression inside the logarithm:

* `g(x) = -21x + 16√x`

Let `y = √x` again. Then `g(x) = -21y² + 16y`.

This is a quadratic function in `y`. To find its maximum, we can complete the square or find the vertex.

The vertex of a quadratic `ay² + by + c` is at `y = -b / (2a)`. In our case:

* `y = -16 / (2 * -21) = 16 / 42 = 8 / 21`

Now, substitute back `y = √x`:

* `√x = 8 / 21`
* `x = (8 / 21)² = 64 / 441`

Since `64/441` is within the domain `(0, 256/441)`, this is a valid maximum.

**3. Verify the Maximum**

To ensure this is a maximum, we can take the second derivative of `g(x)` with respect to `x`:

* `g(x) = -21x + 16x^(1/2)`
* `g'(x) = -21 + 8x^(-1/2)`
* `g''(x) = -4x^(-3/2)`

Since `g''(x)` is negative for all `x > 0`, the function `g(x)` is concave down, and the value we found is indeed a maximum.

**Conclusion**

The value of `x` that maximizes `f(x) = log(-20x + 16√x - x)` is:

* `x = 64 / 441`