Let’s carefully solve the two problems shown in the new image step by step:

Problem 1: Solve $\left(3^{\frac{x}{2}}\right) \times \left(3^{\frac{3x}{2}}\right) = 3^9$

Step 1: Use the rule of exponents

The rule of exponents states: $a^m \cdot a^n = a^{m+n}$ Here, the base is $3$. Combine the exponents: $3^{\frac{x}{2}} \cdot 3^{\frac{3x}{2}} = 3^{\frac{x}{2} + \frac{3x}{2}}$

Add the exponents: $\frac{x}{2} + \frac{3x}{2} = \frac{4x}{2} = 2x$

The left-hand side becomes: $3^{2x}$

The equation now looks like: $3^{2x} = 3^9$

Step 2: Set the exponents equal to each other

Since the bases are the same, equate the exponents: $2x = 9$

Step 3: Solve for $x$

$x = \frac{9}{2} \quad \text{or } 4.5$

Solution to Problem 1:

$x = \frac{9}{2} \quad \text{or } 4.5$

Problem 2: Solve $9^{x-4} = 3^{x-6}$

Step 1: Rewrite the base 9 as $3^2$

The base $9$ can be rewritten as $3^2$: $9^{x-4} = \left(3^2\right)^{x-4} = 3^{2(x-4)} = 3^{2x - 8}$

The equation now becomes: $3^{2x - 8} = 3^{x-6}$

Step 2: Set the exponents equal to each other

Since the bases are the same, equate the exponents: $2x - 8 = x - 6$

Step 3: Solve for $x$

Simplify: $2x - x = -6 + 8$ $x = 2$

Solution to Problem 2:

$x = 2$

Final Answers:

1. For $\left(3^{\frac{x}{2}}\right) \times \left(3^{\frac{3x}{2}}\right) = 3^9$, the solution is: $x = \frac{9}{2} \quad \text{or } 4.5$

2. For $9^{x-4} = 3^{x-6}$, the solution is: $x = 2$

Let me know if you’d like additional clarifications or steps!