If `5^(3x^(2)"log"_(10)2) = 2^((x + (1)/(2))"log"_(10) 25)`, then x equals to

Options

A. `1, -(1)/(3)`
B. 1
C. `1, -(1)/(2)`
D. `-(1)/(3), 1`

Answer

To solve the equation \( 5^{3x^2 \log_{10} 2} = 2^{(x + \frac{1}{2}) \log_{10} 25} \), we can follow these steps:

### Step 1: Rewrite the Right-Hand Side
We know that \( \log_{10} 25 = \log_{10} (5^2) = 2 \log_{10} 5 \). Therefore, we can rewrite the right-hand side as:
\[ 2^{(x + \frac{1}{2}) \cdot 2 \log_{10} 5} = 2^{(2x + 1) \log_{10} 5} \]

### Step 2: Apply the Logarithmic Identity
Using the property \( a^{\log_b x} = x^{\log_b a} \), we can rewrite both sides:
\[ 5^{3x^2 \log_{10} 2} = (2^{(2x + 1)})^{\log_{10} 5} \]
This gives us:
\[ 5^{3x^2 \log_{10} 2} = 5^{(2x + 1) \log_{10} 2} \]

### Step 3: Equate the Exponents
Since the bases are the same, we can set the exponents equal to each other:
\[ 3x^2 \log_{10} 2 = (2x + 1) \log_{10} 2 \]

### Step 4: Simplify the Equation
Assuming \( \log_{10} 2 \neq 0 \) (which is true), we can divide both sides by \( \log_{10} 2 \):
\[ 3x^2 = 2x + 1 \]

### Step 5: Rearrange into Standard Quadratic Form
Rearranging gives us:
\[ 3x^2 - 2x - 1 = 0 \]

### Step 6: Solve the Quadratic Equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = -2, c = -1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ x = \frac{2 \pm \sqrt{4 + 12}}{6} \]
\[ x = \frac{2 \pm \sqrt{16}}{6} \]
\[ x = \frac{2 \pm 4}{6} \]

### Step 7: Calculate the Roots
Calculating the two possible values for \( x \):
1. \( x = \frac{6}{6} = 1 \)
2. \( x = \frac{-2}{6} = -\frac{1}{3} \)

### Step 8: Verify the Solutions
We need to check if both values satisfy the original equation.

1. For \( x = 1 \):
\[ LHS = 5^{3 \cdot 1^2 \log_{10} 2} = 5^{3 \log_{10} 2} = (2^{3})^{\log_{10} 5} = 2^{3 \log_{10} 5} \]
\[ RHS = 2^{(1 + \frac{1}{2}) \log_{10} 25} = 2^{\frac{3}{2} \cdot 2 \log_{10} 5} = 2^{3 \log_{10} 5} \]
Both sides are equal.

2. For \( x = -\frac{1}{3} \):
\[ LHS = 5^{3 \left(-\frac{1}{3}\right)^2 \log_{10} 2} = 5^{\log_{10} 2} \]
\[ RHS = 2^{\left(-\frac{1}{3} + \frac{1}{2}\right) \log_{10} 25} = 2^{\frac{1}{6} \cdot 2 \log_{10} 5} = 2^{\frac{1}{3} \log_{10} 5} \]
Both sides are equal.

### Final Answer
Thus, the values of \( x \) that satisfy the equation are:
\[ x = 1 \quad \text{and} \quad x = -\frac{1}{3} \]