Let `agt1` be a real number . If S is the set of real number `x` that are solutions to the equation `a^(2log_2x)=5+4x^(log_2a)`, then

Options

A. 2
B. infinite
C. 0
D. 1

Explanation

To solve the equation \( a^{2 \log_2 x} = 5 + 4 x^{\log_2 a} \), we will follow these steps:

### Step 1: Rewrite the equation using logarithmic properties
We start with the equation:
\[ a^{2 \log_2 x} = 5 + 4 x^{\log_2 a} \]
Using the property of logarithms, \( a^{\log_b m} = m^{\log_b a} \), we can rewrite \( a^{2 \log_2 x} \) as:
\[ (a^{\log_2 x})^2 = (x^{\log_2 a})^2 \]
Thus, we can express the left side as:
\[ (x^{\log_2 a})^2 = 5 + 4 x^{\log_2 a} \]

### Step 2: Substitute \( t = x^{\log_2 a} \)
Let \( t = x^{\log_2 a} \). Then, the equation becomes:
\[ t^2 = 5 + 4t \]
Rearranging gives us:
\[ t^2 - 4t - 5 = 0 \]

### Step 3: Solve the quadratic equation
Now we can solve the quadratic equation \( t^2 - 4t - 5 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -4, c = -5 \):
\[ t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \]
\[ t = \frac{4 \pm \sqrt{16 + 20}}{2} \]
\[ t = \frac{4 \pm \sqrt{36}}{2} \]
\[ t = \frac{4 \pm 6}{2} \]
This gives us two solutions:
\[ t = \frac{10}{2} = 5 \quad \text{and} \quad t = \frac{-2}{2} = -1 \]

### Step 4: Evaluate the solutions for \( t \)
Since \( t = x^{\log_2 a} \) and \( a > 1 \), \( t \) must be non-negative. Therefore, we discard \( t = -1 \) and keep:
\[ t = 5 \]

### Step 5: Solve for \( x \)
Now we substitute back to find \( x \):
\[ x^{\log_2 a} = 5 \]
Taking logarithm base \( 2 \) on both sides gives:
\[ \log_2 x \cdot \log_2 a = \log_2 5 \]
Thus,
\[ \log_2 x = \frac{\log_2 5}{\log_2 a} \]
Exponentiating both sides, we find:
\[ x = 2^{\frac{\log_2 5}{\log_2 a}} = 5^{\log_2 2} = 5 \]

### Conclusion
The only solution for \( x \) is \( 5 \). Therefore, the number of solutions in the set \( S \) is:
\[ \text{Number of solutions} = 1 \]