If `2^(a)=5, 5^(b)=8, 8^(c )=11 and 11^(d)=14`, then the value of `2^(abcd)` is :

Options

A. 1
B. 2
C. 7
D. 14

Answer

To solve the problem, we need to find the value of \( 2^{abcd} \) given the relationships:

1. \( 2^a = 5 \)
2. \( 5^b = 8 \)
3. \( 8^c = 11 \)
4. \( 11^d = 14 \)

### Step 1: Express \( a \) in terms of logarithms
From the first equation, we can express \( a \) as:
\[ a = \log_2(5) \]

**Hint:** Use the definition of logarithms to express the exponent in terms of the base.

### Step 2: Express \( b \) in terms of logarithms
From the second equation, we can express \( b \) as:
\[ b = \log_5(8) \]

**Hint:** Again, use the definition of logarithms to express \( b \).

### Step 3: Express \( c \) in terms of logarithms
From the third equation, we can express \( c \) as:
\[ c = \log_8(11) \]

**Hint:** Use the logarithmic identity to express \( c \).

### Step 4: Express \( d \) in terms of logarithms
From the fourth equation, we can express \( d \) as:
\[ d = \log_{11}(14) \]

**Hint:** Continue using the logarithmic definition to express \( d \).

### Step 5: Find \( abcd \)
Now we can find \( abcd \):
\[ abcd = \log_2(5) \cdot \log_5(8) \cdot \log_8(11) \cdot \log_{11}(14) \]

Using the change of base formula, we can simplify this:
\[ abcd = \frac{\log(5)}{\log(2)} \cdot \frac{\log(8)}{\log(5)} \cdot \frac{\log(11)}{\log(8)} \cdot \frac{\log(14)}{\log(11)} \]

Notice that the terms will cancel out:
\[ abcd = \frac{\log(14)}{\log(2)} \]

**Hint:** Use the property of logarithms that allows cancellation of terms.

### Step 6: Calculate \( 2^{abcd} \)
Now, we substitute \( abcd \) back into \( 2^{abcd} \):
\[ 2^{abcd} = 2^{\frac{\log(14)}{\log(2)}} \]

Using the property of exponents:
\[ 2^{abcd} = 14 \]

### Final Answer
Thus, the value of \( 2^{abcd} \) is:
\[ \boxed{14} \]