The value of `"log"_(2)"log"_(2)"log"_(4) 256 + 2 "log"_(sqrt(2))2`, is

Options

A. 2
B. 3
C. 5
D. 7

Answer

To solve the expression \( \log_{2} \log_{2} \log_{4} 256 + 2 \log_{\sqrt{2}} 2 \), we will break it down step by step.

### Step 1: Simplify \( \log_{4} 256 \)

First, we know that \( 256 = 4^4 \) because \( 4^4 = (2^2)^4 = 2^8 = 256 \). Therefore, we can write:

\[ \log_{4} 256 = \log_{4} (4^4) = 4 \]

**Hint:** Remember that \( \log_{b} (b^n) = n \).

### Step 2: Substitute back into the expression

Now we substitute \( \log_{4} 256 \) back into the original expression:

\[ \log_{2} \log_{2} 4 + 2 \log_{\sqrt{2}} 2 \]

**Hint:** Keep track of what you substitute to avoid confusion.

### Step 3: Simplify \( \log_{2} 4 \)

Next, we simplify \( \log_{2} 4 \). Since \( 4 = 2^2 \):

\[ \log_{2} 4 = \log_{2} (2^2) = 2 \]

**Hint:** Use the property \( \log_{b} (a^n) = n \log_{b} a \).

### Step 4: Substitute again

Now substitute \( \log_{2} 4 \) back into the expression:

\[ \log_{2} 2 + 2 \log_{\sqrt{2}} 2 \]

**Hint:** Keep substituting step by step to simplify the expression.

### Step 5: Simplify \( \log_{2} 2 \)

We know that:

\[ \log_{2} 2 = 1 \]

**Hint:** Remember that \( \log_{b} b = 1 \).

### Step 6: Simplify \( \log_{\sqrt{2}} 2 \)

Next, we simplify \( \log_{\sqrt{2}} 2 \). Since \( \sqrt{2} = 2^{1/2} \):

\[ \log_{\sqrt{2}} 2 = \frac{\log_{2} 2}{\log_{2} \sqrt{2}} = \frac{1}{1/2} = 2 \]

**Hint:** Use the change of base formula if needed.

### Step 7: Substitute back into the expression

Now substitute \( \log_{\sqrt{2}} 2 \) back into the expression:

\[ 1 + 2 \cdot 2 \]

### Step 8: Calculate the final expression

Now we calculate:

\[ 1 + 4 = 5 \]

Thus, the value of the expression \( \log_{2} \log_{2} \log_{4} 256 + 2 \log_{\sqrt{2}} 2 \) is \( 5 \).

**Final Answer:** 5