The minimum value of `2^((log_(6)3) cos^(2)x ) + 3^((log_(6)2)sin^(2)x)` is

Options

A. `2^(log_(6) sqrt(2)) `
B. `3^(log_(6) sqrt(2))`
C. `2^(1 + log_(6) sqrt(3))`
D. `2 ^(1 - log_(6) sqrt(3))`

Answer

To find the minimum value of the expression \(2^{(\log_{6}3) \cos^{2} x} + 3^{(\log_{6}2) \sin^{2} x}\), we can proceed step by step as follows:

### Step 1: Define the expression
Let:
\[ f(x) = 2^{(\log_{6}3) \cos^{2} x} + 3^{(\log_{6}2) \sin^{2} x} \]

### Step 2: Apply the AM-GM Inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we can state that:
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
for any non-negative \(a\) and \(b\). Here, let:
\[ a = 2^{(\log_{6}3) \cos^{2} x} \quad \text{and} \quad b = 3^{(\log_{6}2) \sin^{2} x} \]

Thus, we have:
\[ f(x) \geq 2 \sqrt{a \cdot b} \]

### Step 3: Calculate \(ab\)
Now, we calculate \(ab\):
\[ ab = 2^{(\log_{6}3) \cos^{2} x} \cdot 3^{(\log_{6}2) \sin^{2} x} \]

Using the property of logarithms, we can rewrite this as:
\[ ab = 2^{(\log_{6}3) \cos^{2} x} \cdot 3^{(\log_{6}2) \sin^{2} x} = (2^{\log_{6}3})^{\cos^{2} x} \cdot (3^{\log_{6}2})^{\sin^{2} x} \]

### Step 4: Simplify using properties of logarithms
Using the change of base formula, we can express:
\[ 2^{\log_{6}3} = 3^{\log_{6}2} \]
Thus, we can substitute:
\[ ab = (3^{\log_{6}2})^{\cos^{2} x} \cdot (3^{\log_{6}2})^{\sin^{2} x} = (3^{\log_{6}2})^{\cos^{2} x + \sin^{2} x} \]
Since \(\cos^{2} x + \sin^{2} x = 1\), we have:
\[ ab = 3^{\log_{6}2} \]

### Step 5: Substitute back into the AM-GM inequality
Now substituting back into the AM-GM inequality:
\[ f(x) \geq 2 \sqrt{3^{\log_{6}2}} = 2 \cdot 3^{\frac{1}{2} \log_{6}2} \]

### Step 6: Simplify further
Using the property of logarithms:
\[ 3^{\frac{1}{2} \log_{6}2} = 2^{\log_{6}3} \]
Thus:
\[ f(x) \geq 2 \cdot 2^{\log_{6}3} = 2^{1 + \log_{6}3} \]

### Step 7: Final result
The minimum value of the expression \(f(x)\) is:
\[ \boxed{2^{1 + \log_{6}3}} \]