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 can follow these steps:

### Step 1: Rewrite the logarithmic equation
We start with the given equation:
\[ 1 + 2 \log_{(2+x)} 5 = \log_5 (x+2) \]

Using the property of logarithms that states \( \log_a b = \frac{1}{\log_b a} \), we can rewrite \( \log_{(2+x)} 5 \) as:
\[ \log_{(2+x)} 5 = \frac{1}{\log_5 (2+x)} \]
Thus, we can rewrite the equation as:
\[ 1 + 2 \cdot \frac{1}{\log_5 (2+x)} = \log_5 (x+2) \]

### Step 2: Substitute and simplify
Let \( t = \log_5 (2+x) \). Then the equation becomes:
\[ 1 + \frac{2}{t} = t \]

### Step 3: Clear the fraction
Multiply through by \( t \) to eliminate the fraction:
\[ t + 2 = t^2 \]
Rearranging gives us:
\[ t^2 - t - 2 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic equation:
\[ (t - 2)(t + 1) = 0 \]
Setting each factor to zero gives us the possible values for \( t \):
\[ t - 2 = 0 \quad \Rightarrow \quad t = 2 \]
\[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \]

### Step 5: Solve for \( x \) using \( t = 2 \)
Substituting \( t = 2 \) back into the equation \( t = \log_5 (2+x) \):
\[ \log_5 (2+x) = 2 \quad \Rightarrow \quad 2+x = 5^2 = 25 \]
Thus,
\[ x = 25 - 2 = 23 \]

### Step 6: Solve for \( x \) using \( t = -1 \)
Now substituting \( t = -1 \):
\[ \log_5 (2+x) = -1 \quad \Rightarrow \quad 2+x = 5^{-1} = \frac{1}{5} \]
Thus,
\[ x = \frac{1}{5} - 2 = \frac{1}{5} - \frac{10}{5} = -\frac{9}{5} \]

### Step 7: Final solutions
The solutions to the equation are:
\[ x = 23 \quad \text{and} \quad x = -\frac{9}{5} \]