Solve` log_(2)(25^(x+3)-1) = 2 + log_(2)(5^(x+3) + 1)`.

Answer

To solve the equation \( \log_{2}(25^{(x+3)} - 1) = 2 + \log_{2}(5^{(x+3)} + 1) \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation
We can rewrite \( 25 \) as \( 5^2 \). Thus, the equation becomes:
\[ \log_{2}((5^{2})^{(x+3)} - 1) = 2 + \log_{2}(5^{(x+3)} + 1) \]
This simplifies to:
\[ \log_{2}(5^{2(x+3)} - 1) = 2 + \log_{2}(5^{(x+3)} + 1) \]

### Step 2: Use logarithmic properties
Using the property \( \log_{a}(b) + \log_{a}(c) = \log_{a}(bc) \), we can rewrite the right side:
\[ \log_{2}(5^{2(x+3)} - 1) = \log_{2}(4) + \log_{2}(5^{(x+3)} + 1) \]
This simplifies to:
\[ \log_{2}(5^{2(x+3)} - 1) = \log_{2}(4(5^{(x+3)} + 1)) \]

### Step 3: Set the arguments equal
Since the logarithms are equal, we can set their arguments equal to each other:
\[ 5^{2(x+3)} - 1 = 4(5^{(x+3)} + 1) \]

### Step 4: Expand and simplify
Expanding the right side:
\[ 5^{2(x+3)} - 1 = 4 \cdot 5^{(x+3)} + 4 \]
Rearranging gives:
\[ 5^{2(x+3)} - 4 \cdot 5^{(x+3)} - 5 = 0 \]

### Step 5: Let \( t = 5^{(x+3)} \)
Let \( t = 5^{(x+3)} \). Then we have:
\[ t^2 - 4t - 5 = 0 \]

### Step 6: Factor the quadratic equation
Factoring the quadratic:
\[ (t - 5)(t + 1) = 0 \]
Thus, the solutions for \( t \) are:
\[ t = 5 \quad \text{or} \quad t = -1 \]

### Step 7: Solve for \( x \)
Since \( t = 5^{(x+3)} \), we consider each case:
1. \( 5^{(x+3)} = 5 \)
- This implies \( x + 3 = 1 \) or \( x = -2 \).

2. \( 5^{(x+3)} = -1 \)
- This case is not possible since \( 5^{(x+3)} \) is always positive.

### Final Solution
Thus, the only solution is:
\[ \boxed{-2} \]

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