If `a gt 1, x gt 0` and `2^(log_(a)(2x))=5^(log_(a)(5x))`, then x is equal to

Options

A. `(1)/(10)`
B. `(1)/(5)`
C. `(1)/(2)`
D. 1

Explanation

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

### Step 1: Take the logarithm of both sides
We start by taking the logarithm of both sides with base \( a \):
\[ \log_a(2^{\log_a(2x)}) = \log_a(5^{\log_a(5x)}) \]

### Step 2: Apply the power rule of logarithms
Using the power rule of logarithms, we can bring down the exponents:
\[ \log_a(2x) \cdot \log_a(2) = \log_a(5x) \cdot \log_a(5) \]

### Step 3: Rewrite the logarithms
We can rewrite \( \log_a(2x) \) and \( \log_a(5x) \):
\[ \log_a(2x) = \log_a(2) + \log_a(x) \quad \text{and} \quad \log_a(5x) = \log_a(5) + \log_a(x) \]
Substituting these into the equation gives:
\[ (\log_a(2) + \log_a(x)) \cdot \log_a(2) = (\log_a(5) + \log_a(x)) \cdot \log_a(5) \]

### Step 4: Expand both sides
Expanding both sides results in:
\[ \log_a(2)^2 + \log_a(2) \cdot \log_a(x) = \log_a(5)^2 + \log_a(5) \cdot \log_a(x) \]

### Step 5: Rearrange the equation
Rearranging the equation to isolate terms involving \( \log_a(x) \):
\[ \log_a(2)^2 - \log_a(5)^2 = \log_a(5) \cdot \log_a(x) - \log_a(2) \cdot \log_a(x) \]
Factoring out \( \log_a(x) \) on the right side gives:
\[ \log_a(2)^2 - \log_a(5)^2 = \log_a(x)(\log_a(5) - \log_a(2)) \]

### Step 6: Solve for \( \log_a(x) \)
Using the difference of squares on the left side:
\[ (\log_a(2) - \log_a(5))(\log_a(2) + \log_a(5)) = \log_a(x)(\log_a(5) - \log_a(2)) \]
Since \( \log_a(5) - \log_a(2) \) is common on both sides, we can simplify:
\[ \log_a(x) = \frac{\log_a(2)^2 - \log_a(5)^2}{\log_a(5) - \log_a(2)} \]

### Step 7: Simplify using the difference of squares
Using the identity \( a^2 - b^2 = (a - b)(a + b) \):
\[ \log_a(x) = \log_a(2) + \log_a(5) \]
This simplifies to:
\[ \log_a(x) = \log_a(10) \]

### Step 8: Exponentiate to solve for \( x \)
Exponentiating both sides gives:
\[ x = 10 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{10} \]