If a, b, c are positive numbers such that `a^(log_(3)7) =27, b^(log_(7)11)=49, c^(log_(11)25)=sqrt(11)`, then the sum of digits of `S=a^((log_(3)7)^(2))+b^((log_(7)11)^(2))+c^((log_(11)25)^(2))` is :

Options

A. 15
B. 17
C. 19
D. 21

Answer

To solve the problem step by step, we start with the given equations and manipulate them to find the required sum \( S \).

### Step 1: Solve for \( a \)
Given:
\[ a^{\log_3 7} = 27 \]
We can express 27 as \( 3^3 \):
\[ a^{\log_3 7} = 3^3 \]
Taking logarithm base 3 on both sides:
\[ \log_3 a \cdot \log_3 7 = 3 \]
Thus,
\[ \log_3 a = \frac{3}{\log_3 7} \]
Now, we can express \( a \) as:
\[ a = 3^{\frac{3}{\log_3 7}} = 3^{3 \cdot \frac{1}{\log_3 7}} = 3^{\log_7 27} \]

### Step 2: Solve for \( b \)
Given:
\[ b^{\log_7 11} = 49 \]
We can express 49 as \( 7^2 \):
\[ b^{\log_7 11} = 7^2 \]
Taking logarithm base 7 on both sides:
\[ \log_7 b \cdot \log_7 11 = 2 \]
Thus,
\[ \log_7 b = \frac{2}{\log_7 11} \]
Now, we can express \( b \) as:
\[ b = 7^{\frac{2}{\log_7 11}} = 7^{2 \cdot \frac{1}{\log_7 11}} = 7^{\log_{11} 49} \]

### Step 3: Solve for \( c \)
Given:
\[ c^{\log_{11} 25} = \sqrt{11} \]
We can express \( \sqrt{11} \) as \( 11^{1/2} \):
\[ c^{\log_{11} 25} = 11^{1/2} \]
Taking logarithm base 11 on both sides:
\[ \log_{11} c \cdot \log_{11} 25 = \frac{1}{2} \]
Thus,
\[ \log_{11} c = \frac{1/2}{\log_{11} 25} \]
Now, we can express \( c \) as:
\[ c = 11^{\frac{1/2}{\log_{11} 25}} = 11^{\frac{1}{2} \cdot \frac{1}{\log_{11} 25}} = 11^{\log_{25} \sqrt{11}} \]

### Step 4: Calculate \( S \)
Now we need to calculate:
\[ S = a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2} \]

Substituting the values we found:
1. For \( a^{(\log_3 7)^2} \):
\[ a^{(\log_3 7)^2} = (3^{\log_7 27})^{(\log_3 7)^2} = 27^{\log_3 7} \]

2. For \( b^{(\log_7 11)^2} \):
\[ b^{(\log_7 11)^2} = (7^{\log_{11} 49})^{(\log_7 11)^2} = 49^{\log_7 11} \]

3. For \( c^{(\log_{11} 25)^2} \):
\[ c^{(\log_{11} 25)^2} = (11^{\log_{25} \sqrt{11}})^{(\log_{11} 25)^2} = \sqrt{11}^{\log_{11} 25} \]

Now substituting these into \( S \):
\[ S = 27^{\log_3 7} + 49^{\log_7 11} + \sqrt{11}^{\log_{11} 25} \]

Using the property \( x^{\log_y z} = z^{\log_y x} \):
\[ S = 7^3 + 11^2 + 25^{1/2} \]
Calculating each term:
- \( 7^3 = 343 \)
- \( 11^2 = 121 \)
- \( 25^{1/2} = 5 \)

Thus,
\[ S = 343 + 121 + 5 = 469 \]

### Final Answer
The sum of the digits of \( S = 469 \) is:
\[ 4 + 6 + 9 = 19 \]