`log_(e)4+log_(4)egt2`
a. True b. False

Answer

To solve the inequality \( \log_e 4 + \log_4 e > 2 \), we will follow these steps:

### Step 1: Rewrite the logarithms using properties of logarithms
We start with the given expression:
\[ \log_e 4 + \log_4 e \]
Using the change of base formula, we can express \( \log_4 e \) in terms of natural logarithms (base \( e \)):
\[ \log_4 e = \frac{\log_e e}{\log_e 4} = \frac{1}{\log_e 4} \]
Thus, we can rewrite the inequality as:
\[ \log_e 4 + \frac{1}{\log_e 4} > 2 \]

### Step 2: Let \( x = \log_e 4 \)
Now, we substitute \( x \) for \( \log_e 4 \):
\[ x + \frac{1}{x} > 2 \]

### Step 3: Multiply through by \( x \) (assuming \( x > 0 \))
Since \( \log_e 4 > 0 \) (because \( 4 > 1 \)), we can safely multiply both sides by \( x \):
\[ x^2 + 1 > 2x \]

### Step 4: Rearrange the inequality
Rearranging gives us:
\[ x^2 - 2x + 1 > 0 \]
This simplifies to:
\[ (x - 1)^2 > 0 \]

### Step 5: Analyze the inequality
The expression \( (x - 1)^2 \) is always non-negative and is equal to zero only when \( x = 1 \). Therefore, \( (x - 1)^2 > 0 \) for all \( x \) except \( x = 1 \).

### Step 6: Determine when \( x = 1 \)
We need to check if \( x = 1 \) corresponds to \( \log_e 4 \):
\[ \log_e 4 = 1 \implies 4 = e^1 \implies e \approx 2.718 < 4 \]
Thus, \( \log_e 4 \neq 1 \).

### Conclusion
Since \( (x - 1)^2 > 0 \) for all \( x \) except \( x = 1 \), and since \( x = \log_e 4 \) is not equal to 1, the inequality holds true for all values of \( x \). Therefore, the original inequality \( \log_e 4 + \log_4 e > 2 \) is **True**.

### Final Answer
a. True

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