91Ó°ÊÓ

Logarithms have several key properties that make them powerful tools for solving equations and inequalities. One crucial property is the **Product Rule**, which states that the logarithm of a product is the sum of the logarithms: \( \log_b(mn) = \log_b(m) + \log_b(n) \). Using this, we can combine logarithms when we see addition inside the logarithmic expressions.
Another useful property is the **Quotient Rule**, providing us with a way to handle subtraction inside logarithms: \( \log_b \left( \frac{m}{n} \right) = \log_b(m) - \log_b(n) \). This allows us to express a difference of logarithms as a single term.

The final property often used is the **Power Rule**: \( \log_b(m^n) = n \log_b(m) \). Though not directly applied here, being familiar with this is helpful in similar contexts. Mastering these properties makes manipulating and solving logarithmic expressions much easier.

Rational inequalities involve expressions with fractions, where the variables appear in the numerator, denominator, or both. When dealing with these inequalities, a key strategy is to make the denominator non-zero. This ensures our manipulations, such as multiplying both sides by the denominator, don't result in undefined terms.

In this exercise, after combining logarithms, we encounter \( \frac{x(x+1)}{2x+6} < 1 \). The approach is to deal with this inequality by multiplying through by the denominator, assuming we keep it non-zero, thus simplifying the process to a polynomial inequality.
This makes it easy to rearrange and solve like traditional inequalities. By handling rational inequalities with care, especially considering the domain restrictions, you can solve them efficiently without errors.

Critical points are specific values that make the numerator or denominator of a rational expression zero. These are vital in analyzing inequalities, as they denote where the sign of an expression might change—essential in establishing solution sets.

For this exercise, we identify the critical points from the expression \((x-3)(x+2) < 0\), which are \(x = 3\) and \(x = -2\). These points highlight where the expression transitions in its sign, thus helping to determine the intervals for testing.
By choosing test points from different intervals defined by these critical points, we ascertain where the inequality holds true. This method ensures you accurately assess which sections of the number line fulfill the inequality condition.

Logarithms are only defined for positive arguments. Hence, when solving logarithmic inequalities, one must always consider the domain restrictions. In simpler terms, any expression inside a logarithm must be greater than zero.

In the exercise provided, the domain condition initial expression was \(\log_2 x + \log_2 (x+1) < \log_2 (2x + 6)\). Here, for the logarithmic functions to be valid, \(x > 0\), \(x+1 > 0\), and \(2x+6 > 0\). This simplifies to \(x > 0\).

Once the initial solution interval of \((-2, 3)\) is found, it's essential to incorporate the domain restriction, ensuring that only values where \(x \in (0, 3)\) are considered, as other values would render the logarithms undefined.