We need to solve three separate problems: 1. Solve the logarithmic equation $\log_2 x + \log_x 4 = 3$.

1. Problem Description

We need to solve three separate problems:

1. Solve the logarithmic equation $\log_2 x + \log_x 4 = 3$.

2. Solve the exponential equation $9^{x+1} - 10\cdot3^x + 1 = 0$.

3. Express the fraction $\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$, where $m$ and $n$ are whole numbers.

2. Solution Steps

Problem 1: $\log_2 x + \log_x 4 = 3$

We can use the change of base formula:

$\log_x 4 = \frac{\log_2 4}{\log_2 x} = \frac{2}{\log_2 x}$.

Let $y = \log_2 x$. Then the equation becomes

$y + \frac{2}{y} = 3$.

Multiplying by $y$, we get

$y^2 + 2 = 3y$.

$y^2 - 3y + 2 = 0$.

$(y-1)(y-2) = 0$.

So, $y=1$ or $y=2$.

If $y=1$, then $\log_2 x = 1$, so $x = 2^1 = 2$.

If $y=2$, then $\log_2 x = 2$, so $x = 2^2 = 4$.

Therefore, $x=2$ or $x=4$.

Problem 2: $9^{x+1} - 10\cdot3^x + 1 = 0$

We can rewrite the equation as

$9^x \cdot 9^1 - 10\cdot3^x + 1 = 0$

$9\cdot (3^x)^2 - 10\cdot3^x + 1 = 0$

Let $y = 3^x$. Then the equation becomes

$9y^2 - 10y + 1 = 0$

$(9y-1)(y-1) = 0$

So, $9y-1=0$ or $y-1=0$.

If $9y-1=0$, then $y = \frac{1}{9}$. Thus, $3^x = \frac{1}{9} = 3^{-2}$, so $x=-2$.

If $y-1=0$, then $y = 1$. Thus, $3^x = 1 = 3^0$, so $x=0$.

Therefore, $x=0$ or $x=-2$.

Problem 3: $\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}}$

We can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{3} + \sqrt{2}$:

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

$= \frac{3\sqrt{2}(2\sqrt{3}) + 3\sqrt{2}(\sqrt{2}) - \sqrt{3}(2\sqrt{3}) - \sqrt{3}(\sqrt{2})}{4(3) - 2} = \frac{6\sqrt{6} + 6 - 6 - \sqrt{6}}{12 - 2} = \frac{5\sqrt{6}}{10} = \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}}$

So $m=6$ and $n=4$.

3. Final Answer

1. $x = 2$ or $x = 4$

2. $x = 0$ or $x = -2$

3. $\frac{\sqrt{6}}{\sqrt{4}}$