6/ The real solutions of the equation `2^(x+2).5^(6-x)=10^(x^2)` is

Options

A. 1
B. 2
C. `-log_(10) (250)`
D. ` log_(10) 4 - 3`

Explanation

To solve the equation \( 2^{(x+2)} \cdot 5^{(6-x)} = 10^{(x^2)} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ 2^{(x+2)} \cdot 5^{(6-x)} = 10^{(x^2)} \]
We can express \( 10^{(x^2)} \) in terms of its prime factors:
\[ 10^{(x^2)} = (2 \cdot 5)^{(x^2)} = 2^{(x^2)} \cdot 5^{(x^2)} \]
Thus, we can rewrite the equation as:
\[ 2^{(x+2)} \cdot 5^{(6-x)} = 2^{(x^2)} \cdot 5^{(x^2)} \]

### Step 2: Equate the powers of 2 and 5
Since the bases are the same, we can equate the exponents:
1. For base 2:
\[ x + 2 = x^2 \]
2. For base 5:
\[ 6 - x = x^2 \]

### Step 3: Solve the equations
#### Equation 1: \( x + 2 = x^2 \)
Rearranging gives:
\[ x^2 - x - 2 = 0 \]
Factoring:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{or} \quad x = -1 \]

#### Equation 2: \( 6 - x = x^2 \)
Rearranging gives:
\[ x^2 + x - 6 = 0 \]
Factoring:
\[ (x - 2)(x + 3) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{or} \quad x = -3 \]

### Step 4: Combine the solutions
From both equations, we have the following solutions:
- From the first equation: \( x = 2, -1 \)
- From the second equation: \( x = 2, -3 \)

The real solutions of the original equation are:
\[ x = 2, -1, -3 \]

### Final Answer
The real solutions of the equation \( 2^{(x+2)} \cdot 5^{(6-x)} = 10^{(x^2)} \) are \( x = 2, -1, -3 \).

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