Rationalize the denominator and simplify. 2. Solve for x in 3 imes extlog3 (x+4) = extlog3 24.

: Rationalize the denominator and simplify

Step 1: Recall the value of $\tan 60^\circ$.

$$\tan 60^\circ = \sqrt{3}$$

Step 2: Substitute the value into the expression.

$$\frac{\sqrt{3}}{\tan 60^\circ - 1} = \frac{\sqrt{3}}{\sqrt{3} - 1}$$

Step 3: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$.

$$\frac{\sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$$

Step 4: Perform the multiplication.

$$\frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{\sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1}{(\sqrt{3})^2 - 1^2}$$ $$= \frac{3 + \sqrt{3}}{3 - 1}$$ $$= \frac{3 + \sqrt{3}}{2}$$

The simplified expression is $\boxed{\frac{3 + \sqrt{3}}{2}}$.

: Solve for x in $3 \log_3 (x+4) = \log_3 24$

Step 1: Use the logarithm property $n \log_b M = \log_b M^n$ on the left side.

$$\log_3 (x+4)^3 = \log_3 24$$

Step 2: Since the bases of the logarithms are the same, equate the arguments.

$$(x+4)^3 = 24$$

Step 3: Take the cube root of both sides.

$$x+4 = \sqrt[3]{24}$$ $$x+4 = \sqrt[3]{8 \times 3}$$ $$x+4 = 2\sqrt[3]{3}$$

Step 4: Solve for x.

$$x = 2\sqrt[3]{3} - 4$$

Step 5: Check the domain restriction for the logarithm. The argument $(x+4)$ must be positive.
Since $2\sqrt[3]{3}$ is approximately $2 \times 1.44 = 2.88$, $x+4 = 2\sqrt[3]{3} \approx 2.88$, which is positive. So the solution is valid.

The value of x is $\boxed{2\sqrt[3]{3} - 4}$.

: Matrix transformation

Step 1: A transformation maps a triangle onto a straight line if the determinant of the transformation matrix is zero.
The given transformation matrix is $M = \begin{bmatrix} x-1 & x \\ 1 & 2x \end{bmatrix}$.

Step 2: Calculate the determinant of the matrix.

$$\det(M) = (x-1)(2x) - (x)(1)$$ $$= 2x^2 - 2x - x$$ $$= 2x^2 - 3x$$

Step 3: Set the determinant to zero and solve for x.

$$2x^2 - 3x = 0$$

Factor out x:

$$x(2x - 3) = 0$$

Step 4: Find the possible values of x.

$$x = 0 \quad \text{or} \quad 2x - 3 = 0$$ $$2x = 3$$ $$x = \frac{3}{2}$$

The possible values of x are $\boxed{0 \text{ or } \frac{3}{2}}$.

: Binomial expansion