Solve the equations:
`3^(2x+1)+3^(2)=3^(x+3)+3^(x).`

Answer

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

### Step 1: Rewrite the equation using properties of exponents
We can rewrite \( 3^{2x+1} \) as \( 3^{2x} \cdot 3^1 \) and \( 3^{x+3} \) as \( 3^x \cdot 3^3 \). Thus, the equation becomes:
\[ 3^{2x} \cdot 3 + 3^2 = 3^x \cdot 27 + 3^x \]
This simplifies to:
\[ 3^{2x+1} + 9 = 27 \cdot 3^x + 3^x \]

### Step 2: Substitute \( 3^x \) with a variable
Let \( t = 3^x \). Then \( 3^{2x} = (3^x)^2 = t^2 \). Substituting these into the equation gives:
\[ 3t^2 + 9 = 28t \]

### Step 3: Rearrange the equation
Rearranging the equation leads to:
\[ 3t^2 - 28t + 9 = 0 \]

### Step 4: Solve the quadratic equation
We can solve the quadratic equation \( 3t^2 - 28t + 9 = 0 \) using the factorization method. We need to find two numbers that multiply to \( 3 \times 9 = 27 \) and add to \( -28 \).

The numbers are \( -27 \) and \( -1 \). Thus, we can rewrite the equation as:
\[ 3t^2 - 27t - t + 9 = 0 \]
Factoring by grouping:
\[ 3t(t - 9) - 1(t - 9) = 0 \]
Factoring out \( (t - 9) \):
\[ (t - 9)(3t - 1) = 0 \]

### Step 5: Find the values of \( t \)
Setting each factor to zero gives:
1. \( t - 9 = 0 \) → \( t = 9 \)
2. \( 3t - 1 = 0 \) → \( t = \frac{1}{3} \)

### Step 6: Substitute back to find \( x \)
Recall that \( t = 3^x \):
1. For \( t = 9 \):
\[ 3^x = 9 \implies 3^x = 3^2 \implies x = 2 \]
2. For \( t = \frac{1}{3} \):
\[ 3^x = \frac{1}{3} \implies 3^x = 3^{-1} \implies x = -1 \]

### Final Answer
Thus, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

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