To express the given quantity as a single logarithm, we will use the logarithmic properties:

• Power Rule: $a \ln(b) = \ln(b^a)$
• Product Rule: $\ln(a) + \ln(b) = \ln(a \cdot b)$
• Quotient Rule: $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$

The expression is:

$\frac{1}{5} \ln((x + 2)^5) + \frac{1}{2} \ln(x) - \ln((x^2 + 3x + 2)^2)$

Step 1: Apply the Power Rule

Apply the power rule to the terms that have coefficients in front of the logarithms.

$\frac{1}{5} \ln((x + 2)^5) = \ln((x + 2))$ $\frac{1}{2} \ln(x) = \ln(x^{1/2}) = \ln(\sqrt{x})$ $\ln((x^2 + 3x + 2)^2) = \ln((x^2 + 3x + 2)^2)$

Now the expression becomes:

$\ln(x + 2) + \ln(\sqrt{x}) - \ln((x^2 + 3x + 2)^2)$

Step 2: Apply the Product Rule

Combine the first two logarithms using the product rule:

$\ln((x + 2) \cdot \sqrt{x})$

So the expression becomes:

$\ln((x + 2) \cdot \sqrt{x}) - \ln((x^2 + 3x + 2)^2)$

Step 3: Apply the Quotient Rule

Now use the quotient rule to combine the remaining terms:

$\ln\left(\frac{(x + 2) \cdot \sqrt{x}}{(x^2 + 3x + 2)^2}\right)$

Thus, the given expression simplifies to:

$\ln\left(\frac{(x + 2) \cdot \sqrt{x}}{(x^2 + 3x + 2)^2}\right)$

This is the simplified expression as a single logarithm.

Would you like further details or have any questions?

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Here are five related questions you might find interesting:

1. How can we expand the expression $\ln((x + 2)^5)$?
2. What is the difference between the product and quotient rules for logarithms?
3. How can the power rule for logarithms simplify calculations with large exponents?
4. Can you derive the expression $\ln\left(\frac{a}{b}\right)$ using the properties of logarithms?
5. What are the common mistakes when combining multiple logarithmic terms?

Tip: Always apply the power rule first when coefficients are present in front of logarithms.