The value of `"lt"_(x to 1)(log_(2)2x)^(log_(x)5)` is:

Options

A. `log_(2)5`
B. `5/2`
C. `e^(log_(2)5)`
D. none

Answer

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

### Step 1: Rewrite the logarithmic expressions
We start by rewriting the logarithmic expressions using properties of logarithms. We know that:
\[ \log_x 5 = \frac{\log_2 5}{\log_2 x} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 1} (\log_2 (2x))^{\frac{\log_2 5}{\log_2 x}} \]

### Step 2: Simplify the logarithm
Next, we simplify \( \log_2 (2x) \):
\[ \log_2 (2x) = \log_2 2 + \log_2 x = 1 + \log_2 x \]
So the limit now looks like:
\[ \lim_{x \to 1} (1 + \log_2 x)^{\frac{\log_2 5}{\log_2 x}} \]

### Step 3: Identify the form
As \( x \to 1 \), \( \log_2 x \to 0 \). Therefore, we have:
\[ 1 + \log_2 x \to 1 \quad \text{and} \quad \frac{\log_2 5}{\log_2 x} \to \infty \]
This gives us the indeterminate form \( 1^\infty \).

### Step 4: Use the exponential limit property
For limits of the form \( 1^\infty \), we can use the property:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) (f(x) - 1)} \]
In our case, let:
\[ f(x) = 1 + \log_2 x \quad \text{and} \quad g(x) = \frac{\log_2 5}{\log_2 x} \]
Thus, we need to evaluate:
\[ \lim_{x \to 1} \frac{\log_2 5}{\log_2 x} \cdot \log_2 x \]

### Step 5: Simplify the limit
This simplifies to:
\[ \lim_{x \to 1} \log_2 5 \cdot \frac{\log_2 x}{\log_2 x} = \lim_{x \to 1} \log_2 5 = \log_2 5 \]

### Step 6: Compute the final limit
Now we can compute the limit:
\[ \lim_{x \to 1} (1 + \log_2 x)^{\frac{\log_2 5}{\log_2 x}} = e^{\log_2 5} = 5 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{5} \]