To rewrite each expression into a single logarithm, we’ll use the laws of logarithms, specifically:

1. Product Rule: $\log_b(M) + \log_b(N) = \log_b(M \cdot N)$
2. Quotient Rule: $\log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right)$
3. Power Rule: $n \cdot \log_b(M) = \log_b(M^n)$

Let's solve each problem.

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(a) $\log_5 (x^2 - 1) - \log_5 (x - 1)$

Using the Quotient Rule:

$\log_5 (x^2 - 1) - \log_5 (x - 1) = \log_5 \left( \frac{x^2 - 1}{x - 1} \right)$

Since $x^2 - 1$ can be factored as $(x + 1)(x - 1)$:

$\log_5 \left( \frac{(x + 1)(x - 1)}{x - 1} \right) = \log_5 (x + 1)$

So, the simplified form is:

$\log_5 (x + 1)$

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(b) $\ln(a + b) + \ln(a - b) - 2 \ln(c)$

First, apply the Product Rule to combine the first two terms:

$\ln(a + b) + \ln(a - b) = \ln((a + b)(a - b))$

Now, use the Power Rule for the last term:

$2 \ln(c) = \ln(c^2)$

Now we have:

$\ln((a + b)(a - b)) - \ln(c^2)$

Finally, apply the Quotient Rule:

$\ln \left( \frac{(a + b)(a - b)}{c^2} \right)$

So, the simplified form is:

$\ln \left( \frac{(a + b)(a - b)}{c^2} \right)$

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(c) $2(\log_5 (x) + 2 \log_5 (y) - 3 \log_5 (z))$

Distribute the $2$ to each term inside the parentheses using the Power Rule:

$2 \log_5 (x) + 4 \log_5 (y) - 6 \log_5 (z)$

Now apply the Power Rule to each term:

$\log_5 (x^2) + \log_5 (y^4) - \log_5 (z^6)$

Next, use the Product Rule for the first two terms and then the Quotient Rule:

$\log_5 \left( x^2 \cdot y^4 \right) - \log_5 (z^6) = \log_5 \left( \frac{x^2 y^4}{z^6} \right)$

So, the simplified form is:

$\log_5 \left( \frac{x^2 y^4}{z^6} \right)$

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(d) $\frac{1}{3} \log((x + 2)^3) + \frac{1}{2} \log(y)$

Use the Power Rule for each term:

$\frac{1}{3} \log((x + 2)^3) = \log((x + 2))$ [ \frac{1}{2} \log(y