Let N = 3/(0.00521).

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

$$\log N = \log \left( \left( \frac{75.4 \times 4.83^2}{0.00521} \right)^{\frac{1}{3}} \right)$$ $$\log N = \frac{1}{3} \left[ \log(75.4) + \log(4.83^2) - \log(0.00521) \right]$$ $$\log N = \frac{1}{3} \left[ \log(75.4) + 2\log(4.83) - \log(0.00521) \right]$$

$$\log(75.4) = 1.8774$$ $$\log(4.83) = 0.6839$$ $$2\log(4.83) = 2 \times 0.6839 = 1.3678$$ $$\log(0.00521) = \bar{3}.7168 \quad (\text{which is } -3 + 0.7168)$$

$$\log N = \frac{1}{3} [1.8774 + 1.3678 - (\bar{3}.7168)]$$ $$\log N = \frac{1}{3} [1.8774 + 1.3678 - (-3 + 0.7168)]$$ $$\log N = \frac{1}{3} [1.8774 + 1.3678 + 3 - 0.7168]$$ $$\log N = \frac{1}{3} [3.2452 + 3 - 0.7168]$$ $$\log N = \frac{1}{3} [6.2452 - 0.7168]$$ $$\log N = \frac{1}{3} [5.5284]$$ $$\log N = 1.8428$$

$$N = \text{antilog}(1.8428)$$

From antilog tables, the antilog of 0.8428 is approximately 6.963.
Since the characteristic is 1, the number has $1+1=2$ digits before the decimal point.

$$N = 69.63$$

The final answer is $\boxed{69.63}$.

---

Add $ad$ to both sides:

$$aNb^2 = bd + ad$$

$$d = \frac{aNb^2}{b + a}$$

The final answer is $\boxed{d = \frac{aNb^2}{b + a}}$.

---

The center is the midpoint of the diameter PQ.
Given points $P(-2, 2)$ and $Q(-2, -6)$.
Let the center be $(h, k)$.

$$h = \frac{x_1 + x_2}{2} = \frac{-2 + (-2)}{2} = \frac{-4}{2} = -2$$ $$k = \frac{y_1 + y_2}{2} = \frac{2 + (-6)}{2} = \frac{-4}{2} = -2$$

The center of the circle is $(-2, -2)$.

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

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

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

Substitute the center $(-2, -2)$ and radius $r=4$:

$$(x - (-2))^2 + (y - (-2))^2 = 4^2$$ $$(x + 2)^2 + (y + 2)^2 = 16$$