We are asked to solve three problems. First, we need to find the value of $\frac{6^{1/2}9^{1/4}}{216^{1/4}}$. Second, we need to solve the equation $\log_2 x + \log_x 4 = 3$. Third, we need to solve the equation $9^{x+1} - 10 \cdot 3^x + 1 = 0$.

1. Problem Description

We are asked to solve three problems. First, we need to find the value of $\frac{6^{1/2}9^{1/4}}{216^{1/4}}$. Second, we need to solve the equation $\log_2 x + \log_x 4 = 3$. Third, we need to solve the equation $9^{x+1} - 10 \cdot 3^x + 1 = 0$.

2. Solution Steps

Problem 9:

We want to find the value of $\frac{6^{1/2}9^{1/4}}{216^{1/4}}$.

$6^{1/2} = (2 \cdot 3)^{1/2} = 2^{1/2} 3^{1/2}$

$9^{1/4} = (3^2)^{1/4} = 3^{2/4} = 3^{1/2}$

$216^{1/4} = (6^3)^{1/4} = 6^{3/4} = (2 \cdot 3)^{3/4} = 2^{3/4} 3^{3/4}$

So, $\frac{6^{1/2}9^{1/4}}{216^{1/4}} = \frac{2^{1/2}3^{1/2}3^{1/2}}{2^{3/4}3^{3/4}} = \frac{2^{1/2}3}{2^{3/4}3^{3/4}} = 2^{1/2 - 3/4} 3^{1 - 3/4} = 2^{-1/4}3^{1/4} = \frac{3^{1/4}}{2^{1/4}} = (\frac{3}{2})^{1/4}$.

Problem 10:

We want to solve $\log_2 x + \log_x 4 = 3$.

We can rewrite $\log_x 4$ as $\frac{\log_2 4}{\log_2 x} = \frac{2}{\log_2 x}$.

Let $y = \log_2 x$. Then the equation becomes $y + \frac{2}{y} = 3$.

Multiplying by $y$ gives $y^2 + 2 = 3y$, or $y^2 - 3y + 2 = 0$.

Factoring gives $(y-1)(y-2) = 0$, 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 $x = 2^2 = 4$.

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

Problem 11:

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

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

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

Factoring gives $(9y - 1)(y - 1) = 0$, so $y = 1$ or $y = \frac{1}{9}$.

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

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

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

3. Final Answer

Problem 9: $(\frac{3}{2})^{1/4}$

Problem 10: 2 or 4

Problem 11: 0 or -2