Solving [log₁₀ 2 + Log₁₀ 2² + ... + Log₁₀ 2ⁿ] = 1 How Many Values Can N Take

Delving into the world of mathematics, we encounter a fascinating problem that intertwines logarithms, the greatest integer function, and the concept of series. The question at hand is: If [log₁₀ 2 + log₁₀ 2² + ... + log₁₀ 2ⁿ] = 1, how many values can n take? This seemingly concise question opens the door to a rich exploration of mathematical principles and problem-solving techniques. To dissect this problem effectively, we need to understand the intricacies of logarithms, the properties of the greatest integer function, and the behavior of series. Let's embark on this mathematical journey, unraveling the layers of the problem to arrive at a comprehensive solution.

Understanding the Core Concepts

Before diving into the solution, it's crucial to solidify our understanding of the fundamental concepts involved. These include logarithms, the greatest integer function, and series. Each of these concepts plays a pivotal role in shaping the problem and guiding us towards the answer.

Logarithms

At its core, a logarithm is the inverse operation to exponentiation. In simpler terms, the logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. Mathematically, if bˣ = y, then log_b y = x, where b is the base. In our problem, we are dealing with base-10 logarithms, denoted as log₁₀. This means we are looking for the power to which 10 must be raised to obtain a specific value. Logarithms possess several key properties that are instrumental in simplifying expressions and solving equations. One such property is the power rule, which states that log_b a^c = c log_b a. This rule will be particularly useful in manipulating the logarithmic terms in our problem.

The Greatest Integer Function

The greatest integer function, often denoted by [x], returns the largest integer less than or equal to x. For instance, [3.14] = 3, [5] = 5, and [-2.7] = -3. This function introduces a discrete element into the problem, as it effectively truncates any decimal portion of a number. Understanding how the greatest integer function behaves is crucial, as it directly impacts the possible values of n that satisfy the given equation. The greatest integer function essentially maps real numbers to integers, and this mapping is not continuous, leading to step-like behavior. This discontinuity is a key characteristic that we must consider when analyzing the problem.

Series

A series is the sum of the terms of a sequence. In our problem, we have a series of logarithmic terms: log₁₀ 2 + log₁₀ 2² + ... + log₁₀ 2ⁿ. To effectively deal with this series, we need to identify any patterns or properties that can help us simplify it. In this case, we can leverage the properties of logarithms to rewrite the series in a more manageable form. Specifically, we can use the power rule of logarithms to bring the exponents outside the logarithmic terms, which will reveal an arithmetic progression. Understanding the behavior of series, particularly arithmetic series, is essential for solving this problem.

Simplifying the Expression

Now that we have a firm grasp of the core concepts, let's apply them to simplify the given expression. The equation we are working with is [log₁₀ 2 + log₁₀ 2² + ... + log₁₀ 2ⁿ] = 1. The first step is to simplify the series of logarithmic terms.

Applying Logarithmic Properties

We can use the power rule of logarithms, log_b a^c = c log_b a, to rewrite each term in the series. This gives us:

log₁₀ 2 + 2 log₁₀ 2 + 3 log₁₀ 2 + ... + n log₁₀ 2

Notice that each term now has a common factor of log₁₀ 2. We can factor this out of the entire series:

log₁₀ 2 (1 + 2 + 3 + ... + n)

The expression inside the parentheses is the sum of the first n natural numbers, which is a well-known arithmetic series. The sum of the first n natural numbers is given by the formula:

1 + 2 + 3 + ... + n = n(n + 1) / 2

Substituting this back into our expression, we get:

log₁₀ 2 * [n(n + 1) / 2]

Now, our original equation becomes:

[log₁₀ 2 * n(n + 1) / 2] = 1

This simplified form is much easier to work with. We have successfully transformed the series of logarithmic terms into a single expression involving n and the constant log₁₀ 2. This simplification is a crucial step in solving the problem.

Utilizing the Greatest Integer Function

With the simplified expression, [log₁₀ 2 * n(n + 1) / 2] = 1, we can now focus on the role of the greatest integer function. The greatest integer function, [x], returns the largest integer less than or equal to x. In our case, the expression inside the greatest integer function, log₁₀ 2 * n(n + 1) / 2, must be such that its greatest integer is equal to 1.

Understanding the Implications

This means that the value of log₁₀ 2 * n(n + 1) / 2 must lie within the interval [1, 2). In other words:

1 ≤ log₁₀ 2 * n(n + 1) / 2 < 2

This inequality provides us with the key to finding the possible values of n. We have effectively translated the problem involving the greatest integer function into a set of inequalities that we can solve. The lower bound of 1 ensures that the greatest integer function will return 1, while the upper bound of 2 ensures that the greatest integer function will not return 2 or any larger integer. This interval is critical for determining the range of possible values for n.

Solving the Inequalities

To solve the inequalities, we can divide all parts by log₁₀ 2:

1 / log₁₀ 2 ≤ n(n + 1) / 2 < 2 / log₁₀ 2

Since log₁₀ 2 ≈ 0.30103, we have:

1 / 0.30103 ≈ 3.3219

2 / 0.30103 ≈ 6.6439

So, our inequalities become:

3. 3219 ≤ n(n + 1) / 2 < 6.6439

Multiplying all parts by 2, we get:

6. 6438 ≤ n(n + 1) < 13.2878

Now, we need to find integer values of n that satisfy this inequality. We can analyze the quadratic expression n(n + 1) for different integer values of n.

Finding the Possible Values of n

We are looking for integer values of n such that 6.6438 ≤ n(n + 1) < 13.2878. Let's test some integer values of n:

• If n = 1, n(n + 1) = 1(2) = 2 (This is less than 6.6438, so n = 1 is not a solution.)
• If n = 2, n(n + 1) = 2(3) = 6 (This is less than 6.6438, so n = 2 is not a solution.)
• If n = 3, n(n + 1) = 3(4) = 12 (This falls within our range, so n = 3 is a solution.)
• If n = 4, n(n + 1) = 4(5) = 20 (This is greater than 13.2878, so n = 4 is not a solution.)

From this analysis, we see that only n = 3 satisfies the inequality. Therefore, there is only one value of n that satisfies the given equation.

Conclusion

In conclusion, if [log₁₀ 2 + log₁₀ 2² + ... + log₁₀ 2ⁿ] = 1, there is only one value that n can take, which is n = 3. This problem beautifully illustrates the interplay between logarithms, the greatest integer function, and series. By carefully applying the properties of logarithms, understanding the behavior of the greatest integer function, and analyzing the resulting inequalities, we were able to arrive at a definitive solution. The key to solving such problems lies in breaking them down into smaller, manageable steps and leveraging the fundamental principles of mathematics.