Which is smaller 2 or `(log_pi 2+log_2 pi)`

Answer

To determine which is smaller between \( 2 \) and \( \log_{\pi} 2 + \log_{2} \pi \), we can follow these steps:

### Step 1: Use the Change of Base Formula
We start by applying the change of base formula for logarithms. The change of base formula states that:
\[ \log_a b = \frac{\log_c b}{\log_c a} \]
Using this, we can rewrite \( \log_{\pi} 2 \) and \( \log_{2} \pi \):
\[ \log_{\pi} 2 = \frac{\log 2}{\log \pi} \quad \text{and} \quad \log_{2} \pi = \frac{\log \pi}{\log 2} \]

### Step 2: Combine the Logarithms
Now we can combine these two logarithmic expressions:
\[ \log_{\pi} 2 + \log_{2} \pi = \frac{\log 2}{\log \pi} + \frac{\log \pi}{\log 2} \]

### Step 3: Find a Common Denominator
To add these fractions, we need a common denominator:
\[ \log_{\pi} 2 + \log_{2} \pi = \frac{(\log 2)^2 + (\log \pi)^2}{\log 2 \cdot \log \pi} \]

### Step 4: Apply the AM-GM Inequality
According to the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
For our case, let \( a = \log 2 \) and \( b = \log \pi \):
\[ \frac{\log 2 + \log \pi}{2} \geq \sqrt{\log 2 \cdot \log \pi} \]
This implies:
\[ \log 2 + \log \pi \geq 2\sqrt{\log 2 \cdot \log \pi} \]

### Step 5: Compare with 2
Now, since we have:
\[ \log_{\pi} 2 + \log_{2} \pi = \frac{(\log 2)^2 + (\log \pi)^2}{\log 2 \cdot \log \pi} \]
We need to show that:
\[ \frac{(\log 2)^2 + (\log \pi)^2}{\log 2 \cdot \log \pi} \geq 2 \]
Multiplying both sides by \( \log 2 \cdot \log \pi \) gives:
\[ (\log 2)^2 + (\log \pi)^2 \geq 2 \log 2 \cdot \log \pi \]
This is true by the AM-GM inequality, indicating that:
\[ \log_{\pi} 2 + \log_{2} \pi \geq 2 \]

### Conclusion
Thus, we conclude that:
\[ 2 < \log_{\pi} 2 + \log_{2} \pi \]
Therefore, \( 2 \) is smaller than \( \log_{\pi} 2 + \log_{2} \pi \).