To express the expression as a single logarithm, let's proceed step-by-step.

Step 1: Start with the expression

$\frac{1}{3} \left[ 2\log(x+3) + \log x - \log(x^2 - 1) \right]$

Step 2: Apply the power rule for logarithms

Using the power rule, $a \log b = \log(b^a)$, we can rewrite $2\log(x+3)$ as $\log((x+3)^2)$: $\frac{1}{3} \left[ \log((x+3)^2) + \log x - \log(x^2 - 1) \right]$

Step 3: Combine terms inside the brackets using the properties of logarithms

Using the product rule, $\log a + \log b = \log(ab)$, and the quotient rule, $\log a - \log b = \log\left(\frac{a}{b}\right)$, we can combine the terms: $\frac{1}{3} \left[ \log\left(\frac{(x+3)^2 \cdot x}{x^2 - 1}\right) \right]$

Step 4: Simplify using the constant outside the brackets

Now, we apply the $\frac{1}{3}$ outside the brackets by using the power rule again: $\log\left(\left(\frac{(x+3)^2 \cdot x}{x^2 - 1}\right)^{\frac{1}{3}}\right)$

Final Answer

$\log\left(\frac{(x+3)^{2} \cdot x}{x^2 - 1}\right)^{\frac{1}{3}}$

Or, alternatively, you could write this as: $\log\left(\frac{(x+3)^{2/3} \cdot x^{1/3}}{(x^2 - 1)^{1/3}}\right)$

Would you like further details, or do you have any questions?

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Here are some related questions that expand on this concept:

1. How does the power rule for logarithms work in other cases with fractional exponents?
2. Can you demonstrate how to simplify a similar expression with trigonometric functions in place of logarithms?
3. What happens if one of the terms inside the logarithm is zero or negative?
4. How can we apply these rules to simplify logarithmic expressions in calculus, especially for derivatives and integrals?
5. Are there cases where combining logs isn't possible, and why?

Tip: Remember that each logarithmic rule (product, quotient, power) helps transform complex expressions into simpler, single-log forms—useful for solving equations efficiently.