The given equation is:

$\log_3^2 x - \log_3 x^3 = \log_3 x^2 - 4$

Let's break this down and solve the equation step by step.

Step 1: Use logarithmic properties.

• The term $\log_3^2 x$ refers to $(\log_3 x)^2$, so it becomes: $(\log_3 x)^2$
• The term $\log_3 x^3$ can be simplified using the power rule of logarithms: $\log_3 x^3 = 3 \log_3 x$
• Similarly, the term $\log_3 x^2$ can also be simplified: $\log_3 x^2 = 2 \log_3 x$

Now, the equation becomes:

$(\log_3 x)^2 - 3 \log_3 x = 2 \log_3 x - 4$

Step 2: Simplify the equation.

Move all terms to one side:

$(\log_3 x)^2 - 3 \log_3 x - 2 \log_3 x + 4 = 0$

Simplify the terms:

$(\log_3 x)^2 - 5 \log_3 x + 4 = 0$

Step 3: Solve the quadratic equation.

Let $y = \log_3 x$. The equation becomes:

$y^2 - 5y + 4 = 0$

Solve this quadratic equation using the factorization method:

$y^2 - 5y + 4 = (y - 4)(y - 1) = 0$

Thus, $y = 4$ or $y = 1$.

Step 4: Back-substitute $y = \log_3 x$.

• If $\log_3 x = 4$, then $x = 3^4 = 81$.
• If $\log_3 x = 1$, then $x = 3^1 = 3$.

Final Solution:

The possible values of $x$ are: $x = 81 \quad \text{or} \quad x = 3$

Would you like more details on any part of the process or have any questions?

Here are 5 related questions for deeper understanding:

1. How do logarithmic identities like the power rule work?
2. Can we solve this equation using a different logarithmic base?
3. What are the real-world applications of logarithmic equations like this?
4. How do we handle logarithmic equations with no real solutions?
5. What happens if we change the base of the logarithms in this equation?

Tip: Always verify that the solutions fit the domain of the original logarithmic equation to avoid extraneous roots.