Determine the value of 'a' which satisfies the equation : `9^(sqrt(a)) + 40^(sqrt(a)) = 41^(sqrt(a))`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( 9^{\sqrt{a}} + 40^{\sqrt{a}} = 41^{\sqrt{a}} \), we can follow these steps:

### Step 1: Recognize the Pythagorean Triplet
We observe that 9, 40, and 41 form a Pythagorean triplet. This means that they satisfy the Pythagorean theorem:
\[ 9^2 + 40^2 = 41^2 \]
Calculating the squares:
\[ 9^2 = 81, \quad 40^2 = 1600, \quad 41^2 = 1681 \]
Now, check if:
\[ 81 + 1600 = 1681 \]
This is true, confirming that \(9, 40, 41\) is a Pythagorean triplet.

### Step 2: Relate the Equation to the Pythagorean Theorem
Since \(9^{\sqrt{a}} + 40^{\sqrt{a}} = 41^{\sqrt{a}}\) resembles the Pythagorean theorem, we can set:
\[ x = \sqrt{a} \]
Thus, the equation becomes:
\[ 9^x + 40^x = 41^x \]

### Step 3: Set \(x\) to 2
From the Pythagorean relationship, we can deduce that if \(x = 2\), then:
\[ 9^2 + 40^2 = 41^2 \]
This means:
\[ \sqrt{a} = 2 \]

### Step 4: Solve for \(a\)
Now, we square both sides to find \(a\):
\[ a = (\sqrt{a})^2 = 2^2 = 4 \]

### Conclusion
Thus, the value of \(a\) that satisfies the original equation is:
\[ \boxed{4} \]