The problem asks for the value of $x$ in the equation:

$\log_x \left( \frac{1}{27} \right) = -3$

This is a logarithmic equation, and it can be solved by converting the logarithmic form into an exponential form. Recall that if:

$\log_b(a) = c$

then:

$b^c = a$

Step 1: Apply the logarithmic rule

We rewrite the given equation:

$\log_x \left( \frac{1}{27} \right) = -3$

in exponential form:

$x^{-3} = \frac{1}{27}$

Step 2: Solve for $x$

The right-hand side of the equation $\frac{1}{27}$ can be rewritten as $27^{-1}$, so:

$x^{-3} = 27^{-1}$

Since the exponents on both sides are negative, we can equate the bases:

$x^3 = 27$

Now solve for $x$:

$x = \sqrt[3]{27} = 3$

Thus, the value of $x$ is $\boxed{3}$.

Correct answer:

a) 3

Would you like a further explanation or more details?

Here are five related questions to explore this topic:

1. How can you express logarithmic equations in exponential form?
2. What are the properties of logarithms that apply when solving equations like this?
3. How do you solve logarithmic equations with negative exponents?
4. What is the relationship between exponents and logarithms?
5. Can you solve logarithmic equations with different bases?

Tip: When solving logarithmic equations, always consider converting them to exponential form if it makes solving easier.