If `log_(2)(a+b)+log_(2)(c+d) ge4`, then the minimum value of a+b+c+d is

Options

A. `2`
B. `4`
C. `6`
D. `8`

Answer

To solve the inequality \( \log_2(a+b) + \log_2(c+d) \geq 4 \) and find the minimum value of \( a+b+c+d \), we can follow these steps:

### Step 1: Combine the logarithms
Using the property of logarithms that states \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \), we can rewrite the inequality:

\[ \log_2((a+b)(c+d)) \geq 4 \]

### Step 2: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides with base 2:

\[ (a+b)(c+d) \geq 2^4 \]

This simplifies to:

\[ (a+b)(c+d) \geq 16 \]

### Step 3: Apply the AM-GM inequality
To find the minimum value of \( a+b+c+d \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. According to AM-GM, for any non-negative numbers \( x \) and \( y \):

\[ \frac{x+y}{2} \geq \sqrt{xy} \]

Let \( x = a+b \) and \( y = c+d \). Then we have:

\[ \frac{(a+b) + (c+d)}{2} \geq \sqrt{(a+b)(c+d)} \]

### Step 4: Substitute the inequality
From the previous step, we know that \( (a+b)(c+d) \geq 16 \). Therefore:

\[ \sqrt{(a+b)(c+d)} \geq \sqrt{16} = 4 \]

This gives us:

\[ \frac{(a+b) + (c+d)}{2} \geq 4 \]

### Step 5: Solve for \( a+b+c+d \)
Multiplying both sides by 2 yields:

\[ (a+b) + (c+d) \geq 8 \]

Thus, we find:

\[ a+b+c+d \geq 8 \]

### Conclusion
The minimum value of \( a+b+c+d \) is **8**.

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