To expand the expression $\log_6 \left( 7a \sqrt[3]{b} \right)$ using logarithmic properties, follow these steps:

Step-by-Step Solution

1. Apply the Product Rule of Logarithms: $\log_b(xy) = \log_b(x) + \log_b(y)$ Here, split $7a \sqrt[3]{b}$ into its factors $7$, $a$, and $\sqrt[3]{b}$: [ \log_6 \left( 7a \sqrt[3]{b} \right) = \log_6(7) + \log_6(a) + \log_6\left( \sqrt[3]{b} \right) ]

2. Simplify the Cube Root Term: Recall that $\sqrt[3]{b}$ is the same as $b^{1/3}$. Using the power rule of logarithms: $\log_b(x^r) = r \log_b(x)$ Apply this to $\log_6\left( b^{1/3} \right)$: $\log_6\left( b^{1/3} \right) = \frac{1}{3} \log_6(b)$

3. Combine the Results: Substitute back into the expanded form: $\log_6 \left( 7a \sqrt[3]{b} \right) = \log_6(7) + \log_6(a) + \frac{1}{3} \log_6(b)$

Final Answer:

$\log_6 \left( 7a \sqrt[3]{b} \right) = \log_6(7) + \log_6(a) + \frac{1}{3} \log_6(b)$

Let me know if you need further clarification or have additional questions!

Related Questions:

1. How do you simplify logarithms with square roots?
2. What is the change of base formula for logarithms?
3. How do you expand $\log_6(x^2y^3)$?
4. Can you explain the properties of logarithms in general?
5. How do you solve logarithmic equations like $\log_6(7a) = 2$?

Tip:

Always remember the key properties of logarithms: product rule, quotient rule, and power rule. They help simplify complex expressions efficiently!