A rectangular plot of land has an area 14m² if the width is (4-√2)m, find the length of the land. 2. Simplify log₁₂6 + log₃6 =

The area of a rectangle is given by $A = L \times W$, where $A$ is the area, $L$ is the length, and $W$ is the width.  
We are given $A = 14 \text{ m}^2$ and $W = (4 - \sqrt{2})\text{ m}$.  
To find the length $L$, we rearrange the formula:
$$
L = \frac{A}{W}
$$
$$
L = \frac{14}{4 - \sqrt{2}}
$$

The conjugate of $(4 - \sqrt{2})$ is $(4 + \sqrt{2})$.
$$
L = \frac{14}{4 - \sqrt{2}} \times \frac{4 + \sqrt{2}}{4 + \sqrt{2}}
$$
$$
L = \frac{14(4 + \sqrt{2})}{(4 - \sqrt{2})(4 + \sqrt{2})}
$$

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ for the denominator:
$$
L = \frac{14(4 + \sqrt{2})}{4^2 - (\sqrt{2})^2}
$$
$$
L = \frac{14(4 + \sqrt{2})}{16 - 2}
$$
$$
L = \frac{14(4 + \sqrt{2})}{14}
$$
$$
L = 4 + \sqrt{2}
$$
The length of the land is $\boxed{(4 + \sqrt{2})\text{ m}}$.

The property states that $\log_b x + \log_b y = \log_b (x \times y)$.  
Here, $b=6$, $x=12$, and $y=3$.
$$
\log_{6}^{12} + \log_{6}^{3} = \log_{6}(12 \times 3)
$$

$$
\log_{6}(12 \times 3) = \log_{6}(36)
$$

We need to find the power to which 6 must be raised to get 36.  
Since $6^2 = 36$, then $\log_{6}(36) = 2$.
$$
\log_{6}^{12} + \log_{6}^{3} = \boxed{2}
$$