Let x = sqrt((75.4 x 4.832)/(0.00521)).

To evaluate using logarithms, we take the logarithm of both sides:

$$\log x = \log \left( \frac{75.4 \times 4.83^2}{0.00521} \right)^{\frac{1}{2}}$$

Using logarithm properties, $\log(A^n) = n\log A$ and $\log\left(\frac{A}{B}\right) = \log A - \log B$:

$$\log x = \frac{1}{2} \left[ \log(75.4) + \log(4.83^2) - \log(0.00521) \right]$$

Using $\log(A^n) = n\log A$:

$$\log x = \frac{1}{2} \left[ \log(75.4) + 2\log(4.83) - \log(0.00521) \right]$$

$$\log(75.4) = 1.8774$$ $$\log(4.83) = 0.6839$$ $$2\log(4.83) = 2 \times 0.6839 = 1.3678$$ $$\log(0.00521) = \bar{3}.7168 \quad \text{or} \quad -2.2832$$

$$\log x = \frac{1}{2} \left[ 1.8774 + 1.3678 - (-2.2832) \right]$$ $$\log x = \frac{1}{2} \left[ 1.8774 + 1.3678 + 2.2832 \right]$$ $$\log x = \frac{1}{2} \left[ 5.5284 \right]$$ $$\log x = 2.7642$$

$$x = \text{antilog}(2.7642)$$ $$x = 10^{2.7642}$$ $$x \approx 580.0$$

The value is $\boxed{580.0}$.

---

This forms a quadratic equation in $b$:

$$aNb^2 - bd - ad = 0$$ $$aNb^2 - db - ad = 0$$

This is a quadratic equation of the form $Ax^2 + Bx + C = 0$, where $x=b$, $A=aN$, $B=-d$, and $C=-ad$.

$$b = \frac{-(-d) \pm \sqrt{(-d)^2 - 4(aN)(-ad)}}{2(aN)}$$ $$b = \frac{d \pm \sqrt{d^2 + 4a^2Nd}}{2aN}$$

The formula with $b$ as the subject is $\boxed{b = \frac{d \pm \sqrt{d^2 + 4a^2Nd}}{2aN}}$.

---

The line PQ is the diameter, so the center of the circle is the midpoint of PQ.
Given coordinates $P(-2, 2)$ and $Q(-2, -6)$.
The midpoint formula is $\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$.
Center $(h, k) = \left( \frac{-2+(-2)}{2}, \frac{2+(-6)}{2} \right)$

$$(h, k) = \left( \frac{-4}{2}, \frac{-4}{2} \right)$$ $$(h, k) = (-2, -2)$$

The radius is half the length of the diameter PQ.
The distance formula for the diameter is $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

$$D = \sqrt{(-2 - (-2))^2 + (-6 - 2)^2}$$ $$D = \sqrt{(0)^2 + (-8)^2}$$ $$D = \sqrt{0 + 64}$$ $$D = \sqrt{64}$$ $$D = 8$$

The radius $r = \frac{D}{2} = \frac{8}{2} = 4$.