The given equation is:

$\log_4(x^3) + \log_2(\sqrt{x}) = 8$

Step-by-step Solution:

1. Simplify the logarithms:

   • Use the logarithm power rule: $\log_b(a^n) = n \cdot \log_b(a)$.
   • For $\log_4(x^3)$: $\log_4(x^3) = 3 \cdot \log_4(x)$
   • For $\log_2(\sqrt{x})$, recall that $\sqrt{x} = x^{1/2}$, so: $\log_2(\sqrt{x}) = \frac{1}{2} \cdot \log_2(x)$

   Substituting these back into the equation: $3 \cdot \log_4(x) + \frac{1}{2} \cdot \log_2(x) = 8$

2. Convert $\log_4(x)$ to base 2: Using the base-change formula $\log_4(x) = \frac{\log_2(x)}{\log_2(4)}$, and knowing $\log_2(4) = 2$, we have: $\log_4(x) = \frac{\log_2(x)}{2}$ Substituting this into the equation: $3 \cdot \frac{\log_2(x)}{2} + \frac{1}{2} \cdot \log_2(x) = 8$

3. Simplify the equation: Combine terms with $\log_2(x)$: $\frac{3}{2} \cdot \log_2(x) + \frac{1}{2} \cdot \log_2(x) = 8$ $2 \cdot \log_2(x) = 8$

4. Solve for $\log_2(x)$: Divide through by 2: $\log_2(x) = 4$

5. Solve for $x$: Rewrite in exponential form: $x = 2^4 = 16$

Final Answer:

$x = 16$

Would you like further details or clarifications? Here are some related questions:

1. How does the base-change formula for logarithms work?
2. What are other rules of logarithms that simplify equations?
3. Can you solve a similar equation with different bases?
4. How does exponential growth relate to logarithms?
5. What happens if the base of one logarithm is irrational?

Tip: Always rewrite logarithmic equations in a single base to simplify calculations.