`(x+x^(log_2x))^7` has fourth term as 4480 then x=

Answer

To solve the problem `(x + x^(log_2 x))^7` having its fourth term equal to 4480, we will use the binomial expansion. Let's break down the solution step by step.

### Step 1: Identify the Binomial Expansion
The expression can be expanded using the binomial theorem:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, let \( a = x \) and \( b = x^{\log_2 x} \), and \( n = 7 \).

### Step 2: Find the Fourth Term
The fourth term in the expansion corresponds to \( k = 3 \) (since the first term corresponds to \( k = 0 \)). Therefore, the fourth term \( T_4 \) is given by:
\[ T_4 = \binom{7}{3} x^{7-3} (x^{\log_2 x})^3 \]
This simplifies to:
\[ T_4 = \binom{7}{3} x^4 (x^{\log_2 x})^3 = \binom{7}{3} x^4 x^{3 \log_2 x} = \binom{7}{3} x^{4 + 3 \log_2 x} \]

### Step 3: Calculate \( \binom{7}{3} \)
Now, we need to calculate \( \binom{7}{3} \):
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
Thus, we have:
\[ T_4 = 35 x^{4 + 3 \log_2 x} \]

### Step 4: Set the Fourth Term Equal to 4480
According to the problem, this fourth term is equal to 4480:
\[ 35 x^{4 + 3 \log_2 x} = 4480 \]

### Step 5: Solve for \( x^{4 + 3 \log_2 x} \)
Dividing both sides by 35 gives:
\[ x^{4 + 3 \log_2 x} = \frac{4480}{35} = 128 \]

### Step 6: Express 128 as a Power of 2
We can express 128 as:
\[ 128 = 2^7 \]
Thus, we have:
\[ x^{4 + 3 \log_2 x} = 2^7 \]

### Step 7: Take Logarithm Base 2
Taking logarithm base 2 on both sides, we get:
\[ (4 + 3 \log_2 x) \log_2 x = 7 \]

### Step 8: Let \( \log_2 x = y \)
Let \( y = \log_2 x \). Then the equation becomes:
\[ 4y + 3y^2 = 7 \]
Rearranging gives:
\[ 3y^2 + 4y - 7 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-7)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{-4 \pm \sqrt{16 + 84}}{6} = \frac{-4 \pm \sqrt{100}}{6} = \frac{-4 \pm 10}{6} \]
This gives us two solutions:
\[ y = 1 \quad \text{and} \quad y = -\frac{7}{3} \]

### Step 10: Find \( x \)
Since \( y = \log_2 x \), we have:
1. If \( y = 1 \), then \( x = 2^1 = 2 \).
2. If \( y = -\frac{7}{3} \), then \( x = 2^{-\frac{7}{3}} \) (not a valid solution since we need positive \( x \)).

Thus, the only valid solution is:
\[ \boxed{2} \]