Solve : `1+ 2 log_(2+x)5=log_5(x+2)`

Answer

To solve the equation \(1 + 2 \log_{(2+x)} 5 = \log_5 (x+2)\), we will follow these steps:

### Step 1: Rewrite the logarithm
We know that \(\log_{a} b = \frac{1}{\log_{b} a}\). Therefore, we can rewrite \(\log_{(2+x)} 5\) as:
\[ \log_{(2+x)} 5 = \frac{1}{\log_{5} (2+x)} \]
Thus, the equation becomes:
\[ 1 + 2 \cdot \frac{1}{\log_{5} (2+x)} = \log_5 (x+2) \]

### Step 2: Let \(t = \log_5 (x+2)\)
Assuming \(t = \log_5 (x+2)\), we can rewrite the equation as:
\[ 1 + \frac{2}{t} = t \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 1 + \frac{2}{t} - t = 0 \]
Multiplying through by \(t\) to eliminate the fraction:
\[ t + 2 - t^2 = 0 \]
This simplifies to:
\[ t^2 - t - 2 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic equation:
\[ (t - 2)(t + 1) = 0 \]
Thus, the solutions for \(t\) are:
\[ t = 2 \quad \text{or} \quad t = -1 \]

### Step 5: Solve for \(x\)
Now, we substitute back for \(t\):
1. For \(t = 2\):
\[ \log_5 (x + 2) = 2 \implies x + 2 = 5^2 = 25 \implies x = 25 - 2 = 23 \]

2. For \(t = -1\):
\[ \log_5 (x + 2) = -1 \implies x + 2 = 5^{-1} = \frac{1}{5} \implies x = \frac{1}{5} - 2 = \frac{1 - 10}{5} = -\frac{9}{5} \]

### Final Solutions
Thus, the solutions to the equation are:
\[ x = 23 \quad \text{and} \quad x = -\frac{9}{5} \]