We are asked to solve three problems. Problem 10: Solve the equation $\log_2{x} + \log_x{4} = 3$ for $x$. Problem 11: Solve the equation $9^{x+1} - 10 \cdot 3^x + 1 = 0$ for $x$. Problem 12: Express $\frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{3}-\sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$ where $m$ and $n$ are whole numbers.

1. Problem Description

We are asked to solve three problems.

Problem 10: Solve the equation $\log_2{x} + \log_x{4} = 3$ for $x$.

Problem 11: Solve the equation $9^{x+1} - 10 \cdot 3^x + 1 = 0$ for $x$.

Problem 12: Express $\frac{3\sqrt{2}-\sqrt{3}}{2\sqrt{3}-\sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$ where $m$ and $n$ are whole numbers.

2. Solution Steps

Problem 10:

We have $\log_2{x} + \log_x{4} = 3$.

Using 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 $y^2 + 2 = 3y$, or $y^2 - 3y + 2 = 0$.

Factoring, we have $(y-1)(y-2) = 0$, so $y=1$ or $y=2$.

If $y=1$, then $\log_2{x} = 1$, which means $x = 2^1 = 2$.

If $y=2$, then $\log_2{x} = 2$, which means $x = 2^2 = 4$.

Therefore, the solutions are $x=2$ and $x=4$.

Problem 11:

We have $9^{x+1} - 10 \cdot 3^x + 1 = 0$.

$9^{x+1} = 9^x \cdot 9^1 = 9 \cdot 9^x = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2$.

Let $y = 3^x$. The equation becomes $9y^2 - 10y + 1 = 0$.

Factoring, we have $(9y-1)(y-1) = 0$.

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

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

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

Therefore, the solutions are $x=0$ and $x=-2$.

Problem 12:

We have $\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}}$.

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{3} + \sqrt{2}$.

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

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

3. Final Answer

Problem 10: 2 or 4

Problem 11: 0 or -2

Problem 12: $\frac{\sqrt{6}}{\sqrt{4}}$