Solving [arccos(-3/π) * Log(3/π)(π/4)] / [1 - 2log(log₂(x))(2)] ≥ 0
Introduction
This article delves into solving the inequality [arccos(-3/π) * log₃/π(π/4)] / [1 - 2log(log₂(x))(2)] ≥ 0. This problem combines trigonometric functions, logarithms, and inequalities, making it a comprehensive exercise in mathematical analysis. We will break down the problem step by step, addressing each component to arrive at the solution. Our approach will involve understanding the properties of the arccosine function, logarithmic functions, and how they interact within the given inequality. Let's embark on this mathematical journey and unravel the solution together.
Understanding the Components
Arccosine Function
The arccosine function, denoted as arccos(x), is the inverse of the cosine function. It returns the angle whose cosine is x. The domain of arccos(x) is [-1, 1], and its range is [0, π]. However, in our case, we have arccos(-3/π). Since π is approximately 3.14159, -3/π is approximately -0.9549. This value falls within the domain of the arccosine function. The arccosine function plays a crucial role in various fields such as physics, engineering, and computer graphics. Understanding its properties is essential for solving complex mathematical problems.
The arccos(-3/π) term is crucial. Since -3/π is within the domain [-1, 1], arccos(-3/π) is a valid real number. The value will be in the range [π/2, π] because the argument is negative. To precisely compute this value, one would typically use a calculator, but for the purpose of solving the inequality, it is enough to recognize that this term is a positive constant.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. The expression logₐ(b) represents the power to which 'a' must be raised to equal 'b.' Several properties of logarithms are crucial for solving this inequality. These include the change of base formula, the product rule, the quotient rule, and the power rule. Understanding these rules allows us to manipulate logarithmic expressions and simplify them.
In our inequality, we encounter log₃/π(π/4) and log(log₂(x))(2). The first term, log₃/π(π/4), has a base of 3/π, which is less than 1. This affects the behavior of the logarithm. Since 3/π < 1, the logarithmic function is decreasing. To analyze this term further, we need to consider the value of π/4 relative to 1. Since π ≈ 3.14159, π/4 is less than 1. A decreasing logarithmic function applied to a value less than 1 yields a positive result. Therefore, log₃/π(π/4) is positive.
The second logarithmic term, log(log₂(x))(2), is more complex. It involves a logarithm within a logarithm. To simplify this, we need to analyze the inner logarithm, log₂(x), first. The value of log₂(x) depends on the value of x. This nested structure requires careful consideration of the domains and ranges of each logarithmic function.
The Inequality
The inequality [arccos(-3/π) * log₃/π(π/4)] / [1 - 2log(log₂(x))(2)] ≥ 0 requires us to determine when the expression is non-negative. The numerator and denominator play different roles in determining the sign of the expression. Analyzing the sign of each component separately and then combining them will lead us to the solution. Inequalities are fundamental in mathematics and are used extensively in optimization problems, calculus, and real analysis.
The given inequality is a fraction, so its sign depends on the signs of the numerator and the denominator. For the entire expression to be greater than or equal to zero, either both the numerator and the denominator must be positive or both must be negative. We must also consider the case where the numerator is zero, as this satisfies the inequality. However, the denominator cannot be zero, as this would make the expression undefined.
Analyzing the Numerator
The numerator is arccos(-3/π) * log₃/π(π/4). We've already established that arccos(-3/π) is positive because -3/π is within the domain of the arccosine function and the result falls in the range [π/2, π]. We also determined that log₃/π(π/4) is positive because the base 3/π is less than 1, and π/4 is also less than 1. Therefore, the product of these two positive terms is positive. This simplifies our problem significantly.
Since both arccos(-3/π) and log₃/π(π/4) are positive, their product is positive. Therefore, the numerator is positive. This means that for the entire inequality to hold, the denominator must also be positive. This simplifies the problem to analyzing the denominator.
Analyzing the Denominator
The denominator is 1 - 2log(log₂(x))(2). For the inequality to hold, this expression must be positive. This gives us the inequality:
1 - 2log(log₂(x))(2) > 0
We need to solve this inequality to find the values of x that satisfy it. This involves several steps, including isolating the logarithmic term and understanding the domain of the logarithmic function.
Solving the Denominator Inequality
First, let's rearrange the inequality:
2log(log₂(x))(2) < 1
log(log₂(x))(2) < 1/2
Now, let's rewrite this in exponential form. Assuming the base of the outer logarithm is 2:
log₂(x) < 2^(1/2)
log₂(x) < √2
Again, rewrite in exponential form:
x < 2^(√2)
This gives us an upper bound for x. However, we also need to consider the domain of the logarithms.
Domain Considerations
We have two logarithmic terms involving x: log₂(x) and log(log₂(x))(2). For log₂(x) to be defined, x must be greater than 0:
x > 0
For log(log₂(x))(2) to be defined, log₂(x) must be greater than 0:
log₂(x) > 0
This means x > 2⁰, so x > 1.
Finally, we need to ensure that log(log₂(x))(2) is defined and the base is positive and not equal to 1. The argument of the outer logarithm is 2, which is positive. The base is log₂(x), so we need log₂(x) > 0 and log₂(x) ≠ 1. We already have log₂(x) > 0, which means x > 1. For log₂(x) ≠ 1, x ≠ 2. Now, let's examine the case where log₂(x) is the base of the logarithm. For the base to be valid, it must be positive and not equal to 1.
- log₂(x) > 0 implies x > 1
- log₂(x) ≠ 1 implies x ≠ 2
Combining these domain restrictions, we have x > 1 and x ≠ 2.
Combining the Results
We have three conditions for x:
- x < 2^(√2)
- x > 1
- x ≠ 2
Combining these conditions, the solution to the inequality is 1 < x < 2^(√2) and x ≠ 2. Note that 2^(√2) ≈ 2.665, so the interval is approximately (1, 2.665), excluding 2.
Final Solution
In summary, to solve the inequality [arccos(-3/π) * log₃/π(π/4)] / [1 - 2log(log₂(x))(2)] ≥ 0, we analyzed the numerator and denominator separately. The numerator was found to be positive, leading us to focus on the denominator. By solving the inequality 1 - 2log(log₂(x))(2) > 0 and considering the domain restrictions imposed by the logarithmic functions, we arrived at the solution 1 < x < 2^(√2), x ≠ 2.
Therefore, the solution to the given inequality is the interval (1, 2) ∪ (2, 2^(√2)).
Conclusion
Solving this inequality required a blend of understanding arccosine, logarithmic functions, and inequality properties. By breaking the problem into smaller parts and analyzing each component, we were able to systematically arrive at the solution. This problem serves as an excellent example of how different mathematical concepts can be combined to create challenging and engaging problems. The final solution, (1, 2) ∪ (2, 2^(√2)), represents the range of x values that satisfy the given inequality. This exercise highlights the importance of domain considerations and the step-by-step approach to solving complex mathematical problems.
This methodical approach can be applied to a wide range of mathematical problems, fostering a deeper understanding of mathematical principles and problem-solving strategies. The combination of trigonometric and logarithmic functions in a single problem emphasizes the interconnectedness of different areas of mathematics. By mastering these fundamental concepts, one can tackle increasingly complex mathematical challenges with confidence and precision.
