The equation `log_(5)x+(log_((x^(2)+3))25)^(-1)=log_(25)10` has

Options

A. no solution
B. one solution
C. two solutions
D. four solutions

Answer

To solve the equation \( \log_5 x + \left( \log_{x^2 + 3} 25 \right)^{-1} = \log_{25} 10 \), we will follow a series of steps involving logarithmic properties.

### Step-by-Step Solution:

1. **Rewrite the Inverse Logarithm**:
\[ \log_5 x + \frac{1}{\log_{x^2 + 3} 25} = \log_{25} 10 \]
Using the property \( \frac{1}{\log_a b} = \log_b a \), we can rewrite the equation:
\[ \log_5 x + \log_{25} (x^2 + 3) = \log_{25} 10 \]

**Hint**: Remember that \( \frac{1}{\log_a b} = \log_b a \) is a useful property for manipulating logarithmic expressions.

2. **Combine the Logarithms**:
Now we can combine the logarithms on the left side:
\[ \log_5 x + \log_{25} (x^2 + 3) = \log_{25} 10 \]
We can express \( \log_{25} (x^2 + 3) \) in terms of base 5:
\[ \log_{25} (x^2 + 3) = \frac{1}{2} \log_5 (x^2 + 3) \]
Thus, the equation becomes:
\[ \log_5 x + \frac{1}{2} \log_5 (x^2 + 3) = \log_{25} 10 \]

**Hint**: Converting logarithms to the same base can simplify the equation.

3. **Express \( \log_{25} 10 \)**:
We can also express \( \log_{25} 10 \) in terms of base 5:
\[ \log_{25} 10 = \frac{1}{2} \log_5 10 \]
Therefore, the equation now looks like:
\[ \log_5 x + \frac{1}{2} \log_5 (x^2 + 3) = \frac{1}{2} \log_5 10 \]

**Hint**: Always try to express all logarithmic terms in the same base for easier manipulation.

4. **Multiply Through by 2**:
To eliminate the fraction, multiply the entire equation by 2:
\[ 2 \log_5 x + \log_5 (x^2 + 3) = \log_5 10 \]

**Hint**: Multiplying through by a constant can help simplify the equation.

5. **Combine Logarithms Again**:
Using the property \( \log_a b + \log_a c = \log_a (bc) \):
\[ \log_5 (x^2) + \log_5 (x^2 + 3) = \log_5 10 \]
This simplifies to:
\[ \log_5 (x^2 (x^2 + 3)) = \log_5 10 \]

**Hint**: Combining logarithmic terms can lead to a simpler equation.

6. **Remove the Logarithm**:
By exponentiating both sides, we get:
\[ x^2 (x^2 + 3) = 10 \]
Expanding this gives:
\[ x^4 + 3x^2 - 10 = 0 \]

**Hint**: Remember that if \( \log_a b = \log_a c \), then \( b = c \).

7. **Solve the Polynomial**:
Let \( y = x^2 \). The equation becomes:
\[ y^2 + 3y - 10 = 0 \]
We can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{49}}{2} \]
This simplifies to:
\[ y = \frac{-3 \pm 7}{2} \]
Thus, we have:
\[ y = 2 \quad \text{or} \quad y = -5 \]
Since \( y = x^2 \), we discard \( y = -5 \) (as \( x^2 \) cannot be negative) and take \( y = 2 \):
\[ x^2 = 2 \implies x = \sqrt{2} \quad \text{or} \quad x = -\sqrt{2} \]

**Hint**: Always check for extraneous solutions, especially when dealing with logarithms.

8. **Count the Solutions**:
The polynomial \( x^4 + 3x^2 - 10 = 0 \) is a quartic equation, which can have up to 4 solutions. Here we found two valid solutions for \( x \) (considering both positive and negative roots).

**Final Conclusion**: The original equation has 4 solutions due to the nature of the polynomial.

### Final Answer:
The equation has **4 solutions**.