Let's start with the first question, where you are asked to solve for $x$ from the equation:

$\log_x 27 = 3$

Step-by-step solution:

1. The logarithmic equation can be rewritten in exponential form: $x^3 = 27$

2. Solve for $x$: $x = \sqrt[3]{27}$ $x = 3$

So, the value of $x$ is $3$.

Next, for the inequality:

You are given: $-3 > x - 9 > 2x$

This is a compound inequality. To solve, split the inequality into two parts:

Step 1: Solve the left inequality:

$-3 > x - 9$

Add 9 to both sides: $6 > x$ This gives: $x < 6$

Step 2: Solve the right inequality:

$x - 9 > 2x$

Subtract $x$ from both sides: $-9 > x$

This gives: $x < -9$

Step 3: Combine the two results:

We find $x < -9$ from the second inequality and $x < 6$ from the first inequality. So, the solution is:

$x < -9$

Would you like more details or have any questions? Here are some related questions you might find useful:

1. How do you convert logarithmic equations to exponential form?
2. What is the general process for solving compound inequalities?
3. Can logarithmic bases have negative values?
4. How do you solve logarithmic equations with variables on both sides?
5. What is the solution set for compound inequalities with overlapping conditions?

Tip: Always check if the solutions to compound inequalities are valid for both parts of the inequality!