Let's simplify each of the given logarithmic expressions step-by-step:

(a) $\frac{\log x^5 + 4 \log x}{\log x^3}$

Step 1: Simplify the numerator $\log x^5 + 4 \log x$:

Using the logarithmic power rule $\log x^n = n \log x$: $\log x^5 = 5 \log x$ So the numerator becomes: $5 \log x + 4 \log x = 9 \log x$

Step 2: Simplify the denominator $\log x^3$:

Using the same power rule: $\log x^3 = 3 \log x$

Step 3: Combine the numerator and denominator: $\frac{9 \log x}{3 \log x}$ Cancel out $\log x$ (since $x > 0$ and $x \neq 1$): $\frac{9}{3} = 3$

Final Answer for (a): $3$

(b) $\log_2 x - 3 \log_8 x$

Step 1: Rewrite $\log_8 x$ in terms of base 2:

Using the change of base formula $\log_a b = \frac{\log_c b}{\log_c a}$, we rewrite: $\log_8 x = \frac{\log_2 x}{\log_2 8}$ Since $8 = 2^3$, $\log_2 8 = 3$. Thus: $\log_8 x = \frac{\log_2 x}{3}$

Step 2: Substitute back into the expression: $\log_2 x - 3 \log_8 x = \log_2 x - 3 \left(\frac{\log_2 x}{3}\right)$

Step 3: Simplify: The second term becomes: $3 \cdot \frac{\log_2 x}{3} = \log_2 x$ Thus: $\log_2 x - \log_2 x = 0$

Final Answer for (b): $0$

(c) $\frac{\log_{125} x^3}{\log_{25} \sqrt[3]{x}}$

Step 1: Simplify $\log_{125} x^3$:

Using the logarithmic power rule: $\log_{125} x^3 = 3 \log_{125} x$

Step 2: Simplify $\log_{25} \sqrt[3]{x}$:

Since $\sqrt[3]{x} = x^{1/3}$, we apply the power rule: $\log_{25} \sqrt[3]{x} = \frac{1}{3} \log_{25} x$

Step 3: Combine the numerator and denominator: $\frac{\log_{125} x^3}{\log_{25} \sqrt[3]{x}} = \frac{3 \log_{125} x}{\frac{1}{3} \log_{25} x}$

Simplify the fraction by multiplying numerator and denominator by 3: $\frac{3 \log_{125} x}{\frac{1}{3} \log_{25} x} = \frac{9 \log_{125} x}{\log_{25} x}$

Step 4: Rewrite $\log_{125} x$ and $\log_{25} x$ in terms of base 5:

Using the change of base formula, we know: $\log_{125} x = \frac{\log_5 x}{\log_5 125}, \quad \log_{25} x = \frac{\log_5 x}{\log_5 25}$ Since $125 = 5^3$, $\log_5 125 = 3$. Similarly, since $25 = 5^2$, $\log_5 25 = 2$. Thus: $\log_{125} x = \frac{\log_5 x}{3}, \quad \log_{25} x = \frac{\log_5 x}{2}$

Step 5: Substitute back into the expression: $\frac{9 \log_{125} x}{\log_{25} x} = \frac{9 \cdot \frac{\log_5 x}{3}}{\frac{\log_5 x}{2}}$

Simplify: $\frac{9 \cdot \frac{\log_5 x}{3}}{\frac{\log_5 x}{2}} = \frac{9}{3} \cdot \frac{2}{1} = 3 \cdot 2 = 6$

Final Answer for (c): $6$

Summary of Answers:

(a) $3$
(b) $0$
(c) $6$

Let me know if you need detailed steps for any part or further clarification!

Related Questions:

1. What is the change of base formula, and how is it applied in logarithms?
2. How does the logarithmic power rule simplify expressions with exponents?
3. Why is it valid to cancel out $\log x$ in the numerator and denominator in part (a)?
4. How do you convert logarithms with different bases (e.g., $\log_{125}$) into base 10 or another base?
5. What are the common logarithmic properties used to simplify expressions?

Tip: Always check if a logarithmic term can be expressed in terms of another base, as this often simplifies the calculation process!