Property of logarithms: The property of logarithms that we used in this problem states that if loga(b) ≥ loga(c), then b ≥ c. This property is true for any base of the logarithm. Note that the inequality symbol flips when you move from the logarithmic expression to the exponential expression.
Solving linear inequalities: To solve a linear inequality like ax + b > c, you need to isolate the variable on one side of the inequality. This involves adding or subtracting terms and possibly multiplying or dividing by constants. You need to be careful when multiplying or dividing by a negative number, as this flips the inequality symbol.
Domain restrictions: When solving logarithmic inequalities, you need to make sure that the argument of the logarithm is positive. This means that the expression inside the logarithm cannot be zero or negative. You need to check for domain restrictions and make sure that any values that make the argument zero or negative are not included in the solution.
Check the solution: After solving the inequality, it's always a good idea to check the solution to make sure that it satisfies the original inequality. You can plug in the solution to the inequality and make sure that both sides of the inequality are still true. If not, you may have made a mistake in solving the inequality.
To solve log2(7x-3) ≥ log2(x+12), you can start by using the fact that if loga(b) ≥ loga(c), then b ≥ c:
Using this property, you can rewrite the inequality as:
7x - 3 ≥ x + 12
Simplifying the inequality by subtracting x and adding 3 from both sides, you get:
6x ≥ 15
Dividing both sides of the inequality by 6, you get:
x ≥ 2.5
Therefore, the solution to the inequality is x ≥ 2.5 which is option C.
Note that when you use the property of logarithms to simplify an inequality, you need to make sure that the argument of the logarithm is positive. In this case, both arguments are positive as long as x > 3/7 and x > -12, respectively. However, these conditions are automatically satisfied when x ≥ 2.5, so there is no need to check them separately.