If `(3log_(10)x+19)/(3log_(10)x-1)=2 log_(10)x+1,` find solution of equation.

Answer

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

### Step 1: Substitute \(y\) for \(\log_{10} x\)
Let \(y = \log_{10} x\). Then, we can rewrite the equation as:
\[ \frac{3y + 19}{3y - 1} = 2y + 1 \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ 3y + 19 = (2y + 1)(3y - 1) \]

### Step 3: Expand the right side
Expanding the right-hand side:
\[ 3y + 19 = 2y \cdot 3y - 2y + 1 \cdot 3y - 1 \]
\[ 3y + 19 = 6y^2 - 2y + 3y - 1 \]
\[ 3y + 19 = 6y^2 + y - 1 \]

### Step 4: Rearrange the equation
Now, rearranging the equation gives:
\[ 0 = 6y^2 + y - 1 - 3y - 19 \]
\[ 0 = 6y^2 - 2y - 20 \]

### Step 5: Simplify the equation
Dividing the entire equation by 2:
\[ 0 = 3y^2 - y - 10 \]

### Step 6: Factor the quadratic equation
To factor \(3y^2 - y - 10\), we can rewrite it:
\[ 3y^2 - 6y + 5y - 10 = 0 \]
Grouping gives:
\[ 3y(y - 2) + 5(y - 2) = 0 \]
Factoring out \((y - 2)\):
\[ (3y + 5)(y - 2) = 0 \]

### Step 7: Solve for \(y\)
Setting each factor to zero gives:
1. \(3y + 5 = 0 \Rightarrow y = -\frac{5}{3}\)
2. \(y - 2 = 0 \Rightarrow y = 2\)

### Step 8: Convert back to \(x\)
Recall that \(y = \log_{10} x\):
1. For \(y = 2\):
\[ \log_{10} x = 2 \Rightarrow x = 10^2 = 100 \]
2. For \(y = -\frac{5}{3}\):
\[ \log_{10} x = -\frac{5}{3} \Rightarrow x = 10^{-\frac{5}{3}} = \frac{1}{10^{\frac{5}{3}}} \]

### Final Solution
Thus, the solutions for \(x\) are:
\[ x = 100 \quad \text{and} \quad x = 10^{-\frac{5}{3}} \]

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