`underset(x to 1)lim (log_(2) 2x)^(log_ x^( 5))` is equal to

Options

A. `log_(2)5`
B. `e^(log_(2^(5))`
C. e
D. 0

Answer

To solve the limit problem \( \lim_{x \to 1} (\log_2 (2x))^{\log_x (5)} \), we will follow these steps:

### Step 1: Rewrite the Limit
We start by rewriting the expression inside the limit:
\[ \lim_{x \to 1} (\log_2 (2x))^{\log_x (5)} = \lim_{x \to 1} \left(\frac{\log_2 (2x)}{\log_2 (2)}\right)^{\frac{\log(5)}{\log(x)}} \]

### Step 2: Simplify the Logarithm
Using the properties of logarithms, we can simplify \( \log_2 (2x) \):
\[ \log_2 (2x) = \log_2 (2) + \log_2 (x) = 1 + \log_2 (x) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 1} \left(1 + \log_2 (x)\right)^{\frac{\log(5)}{\log(x)}} \]

### Step 3: Identify the Form
As \( x \to 1 \), \( \log_2 (x) \to 0 \) and \( \log(x) \to 0 \). This gives us the indeterminate form \( 1^{\infty} \).

### Step 4: Apply the Exponential Limit Rule
For limits of the form \( 1^{\infty} \), we can use the following transformation:
\[ L = e^{\lim_{x \to 1} \left( \log_2 (2x) - \log_2 (2) \right) \cdot \frac{\log(5)}{\log(x)}} \]

### Step 5: Evaluate the Limit
Now we need to evaluate:
\[ \lim_{x \to 1} \left( \log_2 (2x) - 1 \right) \cdot \frac{\log(5)}{\log(x)} \]
Substituting \( \log_2 (2x) = 1 + \log_2 (x) \):
\[ \lim_{x \to 1} \log_2 (x) \cdot \frac{\log(5)}{\log(x)} \]

### Step 6: Change of Base
Using the change of base formula, we have:
\[ \log_2 (x) = \frac{\log(x)}{\log(2)} \]
Thus, the limit becomes:
\[ \lim_{x \to 1} \frac{\log(x)}{\log(2)} \cdot \frac{\log(5)}{\log(x)} = \lim_{x \to 1} \frac{\log(5)}{\log(2)} = \frac{\log(5)}{\log(2)} \]

### Step 7: Final Result
Putting it all together, we have:
\[ L = e^{\frac{\log(5)}{\log(2)}} = 5^{\frac{1}{\log(2)}} = 5^{\log_2 (e)} \]

Thus, the limit evaluates to:
\[ \lim_{x \to 1} (\log_2 (2x))^{\log_x (5)} = 5^{\log_2 (e)} \]