The sum of solution of the equation `log_(10)(3x^(2) +12x +19) - log_(10)(3x +4) = 1` is

Answer

To solve the equation \( \log_{10}(3x^2 + 12x + 19) - \log_{10}(3x + 4) = 1 \), we will follow these steps:

### Step 1: Apply the properties of logarithms
Using the property of logarithms that states \( \log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right) \), we can rewrite the equation:

\[ \log_{10}\left(\frac{3x^2 + 12x + 19}{3x + 4}\right) = 1 \]

### Step 2: Eliminate the logarithm
To eliminate the logarithm, we exponentiate both sides with base 10:

\[ \frac{3x^2 + 12x + 19}{3x + 4} = 10 \]

### Step 3: Multiply both sides by \(3x + 4\)
To eliminate the fraction, multiply both sides by \(3x + 4\):

\[ 3x^2 + 12x + 19 = 10(3x + 4) \]

### Step 4: Expand the right side
Expanding the right side gives:

\[ 3x^2 + 12x + 19 = 30x + 40 \]

### Step 5: Rearrange the equation
Rearranging the equation to bring all terms to one side results in:

\[ 3x^2 + 12x + 19 - 30x - 40 = 0 \]

This simplifies to:

\[ 3x^2 - 18x - 21 = 0 \]

### Step 6: Divide the entire equation by 3
To simplify, divide the entire equation by 3:

\[ x^2 - 6x - 7 = 0 \]

### Step 7: Factor the quadratic equation
Now, we will factor the quadratic equation:

\[ (x - 7)(x + 1) = 0 \]

### Step 8: Solve for \(x\)
Setting each factor to zero gives us the solutions:

1. \(x - 7 = 0 \implies x = 7\)
2. \(x + 1 = 0 \implies x = -1\)

### Step 9: Check the validity of the solutions
We need to ensure that both solutions satisfy the original logarithmic conditions (the arguments must be positive):

- For \(x = 7\):
- \(3(7)^2 + 12(7) + 19 = 147 + 84 + 19 = 250\) (positive)
- \(3(7) + 4 = 21 + 4 = 25\) (positive)

- For \(x = -1\):
- \(3(-1)^2 + 12(-1) + 19 = 3 - 12 + 19 = 10\) (positive)
- \(3(-1) + 4 = -3 + 4 = 1\) (positive)

Both solutions are valid.

### Step 10: Find the sum of the solutions
The sum of the solutions is:

\[ 7 + (-1) = 6 \]

### Final Answer
The sum of the solutions of the equation is \( \boxed{6} \).

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