The value of `a^x-b^y` is (where`x=sqrt(log_ab)`and `y=sqrt(log_ba),agt0,bgt0` and `a,bne1`)

Options

A. 1
B. 2
C. 0
D. -1

Answer

To solve the problem, we need to find the value of \( a^x - b^y \) where \( x = \sqrt{\log_a b} \) and \( y = \sqrt{\log_b a} \). Let's break this down step by step.

### Step 1: Write down the expressions for \( x \) and \( y \)
We have:
\[ x = \sqrt{\log_a b} \]
\[ y = \sqrt{\log_b a} \]

### Step 2: Rewrite \( a^x \)
We can express \( a^x \) as:
\[ a^x = a^{\sqrt{\log_a b}} \]
Using the property of logarithms, we can rewrite this as:
\[ a^{\sqrt{\log_a b}} = a^{\sqrt{\frac{\log b}{\log a}}} \]

### Step 3: Simplify \( a^x \)
Using the property \( a^{\log_a b} = b \), we can rewrite:
\[ a^{\sqrt{\log_a b}} = b^{\frac{1}{\sqrt{\log_a b}}} \]
This means:
\[ a^x = b^{\frac{1}{\sqrt{\log_a b}}} \]

### Step 4: Rewrite \( b^y \)
Now, let's express \( b^y \):
\[ b^y = b^{\sqrt{\log_b a}} \]
Using the property of logarithms again, we can rewrite this as:
\[ b^{\sqrt{\log_b a}} = a^{\frac{1}{\sqrt{\log_b a}}} \]

### Step 5: Simplify \( b^y \)
Using the property \( b^{\log_b a} = a \), we can rewrite:
\[ b^{\sqrt{\log_b a}} = a^{\frac{1}{\sqrt{\log_b a}}} \]
This means:
\[ b^y = a^{\frac{1}{\sqrt{\log_b a}}} \]

### Step 6: Find \( a^x - b^y \)
Now we have:
\[ a^x = b^{\frac{1}{\sqrt{\log_a b}}} \]
\[ b^y = a^{\frac{1}{\sqrt{\log_b a}}} \]
Thus, we can find:
\[ a^x - b^y = b^{\frac{1}{\sqrt{\log_a b}}} - a^{\frac{1}{\sqrt{\log_b a}}} \]

### Step 7: Evaluate \( a^x - b^y \)
Since \( \sqrt{\log_a b} \) and \( \sqrt{\log_b a} \) are reciprocals, we find that:
\[ a^x = b^y \]
Thus:
\[ a^x - b^y = 0 \]

### Final Answer
The value of \( a^x - b^y \) is:
\[ \boxed{0} \]