Use logarithms correct to 4 decimal places to evaluate sqrt((0.3698 56)/(2.548)).

Taking the logarithm of both sides:

$$\log N = \log \left( \frac{0.3698 \sin 56}{2.548} \right)^{1/2}$$ $$\log N = \frac{1}{2} \left( \log(0.3698) + \log(\sin 56^\circ) - \log(2.548) \right)$$

Using a calculator for $\sin 56^\circ$:
$\sin 56^\circ \approx 0.8290$ (to 4 decimal places)

Now, find the logarithms:
$\log(0.3698) = \bar{1}.5680$
$\log(\sin 56^\circ) = \log(0.8290) = \bar{1}.9186$
$\log(2.548) = 0.4062$

$$\log N = \frac{1}{2} \left( \bar{1}.5680 + \bar{1}.9186 - 0.4062 \right)$$

First, add the logarithms in the numerator:

$$\bar{1}.5680 + \bar{1}.9186 = (-1 + 0.5680) + (-1 + 0.9186) = -2 + 1.4866 = \bar{1}.4866$$

Now, subtract $\log(2.548)$:

$$\bar{1}.4866 - 0.4062 = (-1 + 0.4866) - 0.4062 = -1 + 0.0804 = \bar{1}.0804$$

So,

$$\log N = \frac{1}{2} (\bar{1}.0804)$$

To divide the characteristic $\bar{1}$ by 2, we rewrite it as $\bar{2} + 1$:

$$\log N = \frac{1}{2} (-2 + 1.0804)$$ $$\log N = -1 + 0.5402 = \bar{1}.5402$$

Antilog($\bar{1}.5402$) = Antilog($-1 + 0.5402$)
Antilog(0.5402) $\approx 3.469$
Therefore, $N \approx 0.3469$.

The value is:

$$\boxed{0.3469}$$

:

Solve the equation below: $7^{2x} - 8 \times 7^x + 7 = 0$.

Let $y = 7^x$. Then $7^{2x} = (7^x)^2 = y^2$.
The equation becomes a quadratic equation in terms of $y$:

$$y^2 - 8y + 7 = 0$$

Factor the quadratic equation:

$$(y-1)(y-7) = 0$$

This gives two possible values for $y$:

$$y = 1 \quad \text{or} \quad y = 7$$

Case 1: $y = 1$

$$7^x = 1$$

Since $1 = 7^0$, we have:

$$7^x = 7^0$$ $$x = 0$$

Case 2: $y = 7$

$$7^x = 7$$

Since $7 = 7^1$, we have:

$$7^x = 7^1$$ $$x = 1$$

The solutions for $x$ are:

$$\boxed{x=0, x=1}$$

:

If $\frac{\sqrt{14}}{\sqrt{7}-\sqrt{2}} + \frac{\sqrt{14}}{\sqrt{7}+\sqrt{2}} = a\sqrt{7}+