Key Concept: The Binomial Theorem

For any binomial expansion of the form $(a+b)^n$, the general term, often denoted as $T_{r+1}$, gives us the $(r+1)$-th term in the expansion. Its formula is: $T_{r+1} = {}^nC_r a^{n-r} b^r$ where:

• $n$ is the power to which the binomial is raised.
• $r$ is the index of the term (starting from $r=0$ for the first term).
• ${}^nC_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, representing the number of ways to choose $r$ items from a set of $n$ items.

In this problem, the given expression is $(x + x^{{\log_2}x})^7$. Comparing this with $(a+b)^n$, we identify:

• $a = x$
• $b = x^{{\log_2}x}$
• $n = 7$

Step 1: Determine the Fourth Term (T₄)

The problem states that the fourth term in the expansion is 4480. For the fourth term, $T_{r+1} = T_4$, which implies $r+1 = 4$, so $r = 3$.

Now, we substitute these values into the general term formula: $T_4 = {}^7C_3 (x)^{7-3} (x^{{\log_2}x})^3$ $T_4 = {}^7C_3 x^4 (x^{{\log_2}x})^3$

Step 2: Calculate the Binomial Coefficient

Let's calculate the value of ${}^7C_3$: ${}^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5 \times 4!}{ (3 \times 2 \times 1) \times 4! } = \frac{7 \times 6 \times 5}{6} = 35$

So, the fourth term can be written as: $T_4 = 35 x^4 (x^{{\log_2}x})^3$

Step 3: Formulate and Simplify the Equation

We are given that the fourth term is 4480. Therefore, we can set up the equation: $35 x^4 (x^{{\log_2}x})^3 = 4480$

To simplify, we first divide both sides by 35: $x^4 (x^{{\log_2}x})^3 = \frac{4480}{35}$ $x^4 (x^{{\log_2}x})^3 = 128$

Now, we use exponent rules to simplify the left side. Recall that $(a^m)^n = a^{mn}$: $(x^{{\log_2}x})^3 = x^{3 \cdot {\log_2}x}$ The equation becomes: $x^4 \cdot x^{3 {\log_2}x} = 128$

Next, use the exponent rule $a^m \cdot a^n = a^{m+n}$: $x^{4 + 3 {\log_2}x} = 128$

We also know that $128 = 2^7$. So the equation is: $x^{4 + 3 {\log_2}x} = 2^7$

Step 4: Solve the Equation Using Logarithms

To solve for $x$ when it appears in both the base and the exponent, taking the logarithm of both sides is a common strategy. Since the logarithm in the exponent is base 2, it is most convenient to take the logarithm base 2 on both sides of the equation.

Recall the logarithm property: $\log_b (M^P) = P \log_b M$. Taking $\log_2$ on both sides: $\log_2 (x^{4 + 3 {\log_2}x}) = \log_2 (2^7)$

Applying the logarithm property to both sides: $(4 + 3 {\log_2}x) \cdot {\log_2}x = 7 \cdot {\log_2}2$ Since ${\log_2}2 = 1$: $(4 + 3 {\log_2}x) {\log_2}x = 7$

To make this equation easier to solve, let's substitute $y = {\log_2}x$. The equation transforms into a quadratic equation in terms of $y$: $(4 + 3y)y = 7$ $4y + 3y^2 = 7$ Rearranging into standard quadratic form $ay^2 + by + c = 0$: $3y^2 + 4y - 7 = 0$

Now we solve this quadratic equation for $y$. We can factor it: We need two numbers that multiply to $3 \times (-7) = -21$ and add up to $4$. These numbers are $7$ and $-3$. $3y^2 + 7y - 3y - 7 = 0$ Group terms and factor: $y(3y + 7) - 1(3y + 7) = 0$ $(y - 1)(3y + 7) = 0$ This gives two possible values for $y$: $y - 1 = 0 \implies y = 1$ $3y + 7 = 0 \implies y = -\frac{7}{3}$

Step 5: Find the Value of x and Apply Conditions

Now, we substitute back $y = {\log_2}x$ for each value of $y$:

Case 1: $y = 1$ ${\log_2}x = 1$ To find $x$, we convert the logarithmic equation to its exponential form: $x = 2^1$ $x = 2$

Case 2: $y = -\frac{7}{3}$ ${\log_2}x = -\frac{7}{3}$ Converting to exponential form: $x = 2^{-7/3}$ $x = \frac{1}{2^{7/3}} = \frac{1}{\sqrt[3]{2^7}} = \frac{1}{\sqrt[3]{128}}$

The problem statement specifies that $x \in N$, which means $x$ must be a natural number (positive integers: 1, 2, 3, ...).

• From Case 1, $x = 2$. This is a natural number.
• From Case 2, $x = 2^{-7/3}$. This is a fraction and not a natural number.

Therefore, the only valid value of $x$ that satisfies the given conditions is $x = 2$.

Tips and Common Mistakes:

• Understanding the General Term: Ensure you correctly identify $a$, $b$, and $n$ from the binomial expression and the correct $r$ for the desired term. A common mistake is using $r=4$ for the fourth term instead of $r=3$.
• Exponent Rules: Be careful when simplifying terms like $(x^{{\log_2}x})^3$. Remember $(A^m)^n = A^{mn}$.
• Logarithm Properties: The key to solving this problem lies in correctly applying $\log_b (M^P) = P \log_b M$ and $\log_b (MN) = \log_b M + \log_b N$. Ensure the base of the logarithm chosen matches the problem context for simplification.
• Domain of x: Always check any conditions given for $x$ (e.g., $x \in N$) to filter out extraneous solutions.

Summary and Key Takeaway:

This problem is an excellent example of how different mathematical concepts—the Binomial Theorem, Exponents, and Logarithms—are integrated into a single question. The solution involves:

1. Using the Binomial Theorem to set up an equation for a specific term.
2. Simplifying the exponential expression using exponent rules.
3. Employing logarithm properties to transform the exponential equation into a more manageable quadratic form.
4. Solving the quadratic equation and then back-substituting to find the value of the original variable.
5. Finally, verifying the solutions against the given domain constraints for $x$. The valid natural number solution for $x$ is 2.