If `log_(x)(3x^(2)+10x)=3` , then x=

Options

A. `0`
B. `-2`
C. `5`
D. none

Answer

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

### Step 1: Rewrite the logarithmic equation in exponential form
The equation \( \log_{x}(3x^{2} + 10x) = 3 \) can be rewritten in exponential form:
\[ 3x^{2} + 10x = x^{3} \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ x^{3} - 3x^{2} - 10x = 0 \]

### Step 3: Factor out the common term
We can factor out \( x \) from the equation:
\[ x(x^{2} - 3x - 10) = 0 \]

### Step 4: Solve for \( x \)
This gives us two cases to solve:
1. \( x = 0 \)
2. \( x^{2} - 3x - 10 = 0 \)

### Step 5: Solve the quadratic equation
We can solve the quadratic equation \( x^{2} - 3x - 10 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]
Here, \( a = 1, b = -3, c = -10 \):
\[ x = \frac{3 \pm \sqrt{(-3)^{2} - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \]
\[ x = \frac{3 \pm \sqrt{9 + 40}}{2} \]
\[ x = \frac{3 \pm \sqrt{49}}{2} \]
\[ x = \frac{3 \pm 7}{2} \]

Calculating the two possible values:
1. \( x = \frac{10}{2} = 5 \)
2. \( x = \frac{-4}{2} = -2 \)

### Step 6: Check the validity of the solutions
We need to check which of these values are valid for the logarithm:
- \( x = 0 \): Not valid (logarithm is undefined for non-positive bases).
- \( x = 5 \): Valid (5 is positive and not equal to 1).
- \( x = -2 \): Not valid (logarithm is undefined for negative bases).

### Final Answer
Thus, the only valid solution is:
\[ \boxed{5} \]