Set the recurring decimal equal to x.

Step 1: Set the recurring decimal equal to $x$.
Let $x = 12.1\dot{8}$.
This means $x = 12.1888...$

Step 2: Multiply by a power of 10 to move the decimal point just before the repeating part.
Multiply by 10:

$$10x = 121.888... \quad (1)$$

Step 3: Multiply by a power of 10 to move the decimal point past one full repeating block.
Since one digit repeats, multiply $x$ by 100 (to move the decimal past the non-repeating '1' and the repeating '8'):

$$100x = 1218.888... \quad (2)$$

Step 4: Subtract equation (1) from equation (2).

$$100x - 10x = 1218.888... - 121.888...$$ $$90x = 1097$$

Step 5: Solve for $x$.

$$x = \frac{1097}{90}$$

The fraction cannot be simplified further.

The recurring decimal $12.1\dot{8}$ as a fraction is $\boxed{\frac{1097}{90}}$.

a) Evaluate $\frac{39.51 \times 614}{0.758}$ using logarithms.

Step 1: Let $Y$ be the expression and take the logarithm base 10 of both sides.

$$Y = \frac{39.51 \times 614}{0.758}$$ $$\log_{10} Y = \log_{10} \left( \frac{39.51 \times 614}{0.758} \right)$$

Using logarithm properties $\log(AB) = \log A + \log B$ and $\log(\frac{A}{B}) = \log A - \log B$:

$$\log_{10} Y = \log_{10}(39.51) + \log_{10}(614) - \log_{10}(0.758)$$

Step 2: Find the logarithm of each number (using 4 decimal places).

$$\log_{10}(39.51) \approx 1.5967$$ $$\log_{10}(614) \approx 2.7882$$ $$\log_{10}(0.758) \approx \bar{1}.8801 \text{ or } -0.1199$$

Step 3: Perform the addition and subtraction.

$$\log_{10} Y = 1.5967 + 2.7882 - (-0.1199)$$ $$\log_{10} Y = 1.5967 + 2.7882 + 0.1199$$ $$\log_{10} Y = 4.5048$$

Step 4: Find the antilogarithm.

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