Prove that `asqrt(log_a b)-bsqrt(log_b a)=0`

Answer

To prove that \( a \sqrt{\log_a b} - b \sqrt{\log_b a} = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the equation:
\[ a \sqrt{\log_a b} = b \sqrt{\log_b a} \]

### Step 2: Square both sides
Next, square both sides to eliminate the square roots:
\[ (a \sqrt{\log_a b})^2 = (b \sqrt{\log_b a})^2 \]
This simplifies to:
\[ a^2 \log_a b = b^2 \log_b a \]

### Step 3: Use the change of base formula
Using the change of base formula, we can express \(\log_a b\) and \(\log_b a\) in terms of natural logarithms:
\[ \log_a b = \frac{\log b}{\log a} \quad \text{and} \quad \log_b a = \frac{\log a}{\log b} \]
Substituting these into the equation gives:
\[ a^2 \cdot \frac{\log b}{\log a} = b^2 \cdot \frac{\log a}{\log b} \]

### Step 4: Cross-multiply
Cross-multiplying yields:
\[ a^2 \log b \cdot \log b = b^2 \log a \cdot \log a \]
This simplifies to:
\[ a^2 (\log b)^2 = b^2 (\log a)^2 \]

### Step 5: Divide both sides
Now, divide both sides by \((\log a)(\log b)\) (assuming \(\log a \neq 0\) and \(\log b \neq 0\)):
\[ \frac{a^2 \log b}{\log a} = \frac{b^2 \log a}{\log b} \]

### Step 6: Rearranging
Rearranging gives:
\[ \frac{a^2}{b^2} = \frac{\log a}{\log b} \cdot \frac{\log b}{\log a} \]
This implies:
\[ \frac{a^2}{b^2} = 1 \]

### Step 7: Taking square roots
Taking square roots of both sides gives:
\[ \frac{a}{b} = 1 \quad \text{or} \quad \frac{a}{b} = -1 \]
Since \(a\) and \(b\) are positive, we have:
\[ a = b \]

### Conclusion
Thus, we have shown that \( a \sqrt{\log_a b} - b \sqrt{\log_b a} = 0 \) holds true under the condition that \( a = b \).

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