To solve the equation $\log_3 (x^3) - 4 \log_9 (x) - 5 \log_{27} (\sqrt{x}) = \log_9 (4)$, we'll start by converting all logarithms to the same base, preferably base 3, and simplifying step by step.

Step 1: Convert logarithms to base 3

We'll use the change of base formula: $\log_b (a) = \frac{\log_c (a)}{\log_c (b)}$ We know that $\log_9 (x) = \frac{1}{2} \log_3 (x)$ since $9 = 3^2$, and similarly, $\log_{27} (x) = \frac{1}{3} \log_3 (x)$ since $27 = 3^3$.

The right-hand side $\log_9 (4)$ can also be written as: $\log_9 (4) = \frac{1}{2} \log_3 (4)$

Thus, the equation becomes: $\log_3 (x^3) - 4 \left( \frac{1}{2} \log_3 (x) \right) - 5 \left( \frac{1}{3} \log_3 (x^{1/2}) \right) = \frac{1}{2} \log_3 (4)$

Step 2: Simplify logarithmic terms

Now let's simplify each term:

• $\log_3 (x^3) = 3 \log_3 (x)$
• The second term becomes $4 \times \frac{1}{2} \log_3 (x) = 2 \log_3 (x)$
• The third term: $\log_3 (x^{1/2}) = \frac{1}{2} \log_3 (x)$, so $5 \times \frac{1}{3} \times \frac{1}{2} \log_3 (x) = \frac{5}{6} \log_3 (x)$

Thus, the equation now is: $3 \log_3 (x) - 2 \log_3 (x) - \frac{5}{6} \log_3 (x) = \frac{1}{2} \log_3 (4)$

Step 3: Combine logarithmic terms

On the left-hand side, combine like terms: $(3 - 2 - \frac{5}{6}) \log_3 (x) = \frac{1}{2} \log_3 (4)$ $\left( \frac{6}{6} - \frac{5}{6} \right) \log_3 (x) = \frac{1}{2} \log_3 (4)$ $\frac{1}{6} \log_3 (x) = \frac{1}{2} \log_3 (4)$

Step 4: Solve for $\log_3 (x)$

Multiply both sides by 6 to isolate $\log_3 (x)$: $\log_3 (x) = 3 \log_3 (4)$ Thus, $x = 4^3 = 64$.

Final Answer:

The solution is $x = 64$.

Would you like more details on any step or another explanation?

Here are some related questions you might consider:

1. How does the change of base formula work for logarithms?
2. What are the properties of logarithms used in simplification?
3. How can we apply logarithms to solve exponential equations?
4. What happens if you solve this equation in terms of a different base, like 10 or e?
5. How do logarithmic identities assist in solving complex logarithmic expressions?

Tip: Always check that the argument of a logarithm is positive, as logarithms are undefined for non-positive numbers.