Solve `3^(1+x)+3^(1-x)=10`.

Answer

To solve the equation \( 3^{1+x} + 3^{1-x} = 10 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3^{1+x} + 3^{1-x} = 10 \]
We can rewrite \( 3^{1+x} \) as \( 3 \cdot 3^x \) and \( 3^{1-x} \) as \( 3 \cdot 3^{-x} \):
\[ 3 \cdot 3^x + 3 \cdot 3^{-x} = 10 \]

### Step 2: Factor out the common term
Now, factor out the 3 from the left side:
\[ 3(3^x + 3^{-x}) = 10 \]
Dividing both sides by 3 gives:
\[ 3^x + 3^{-x} = \frac{10}{3} \]

### Step 3: Substitute \( t \)
Let \( t = 3^x \). Then \( 3^{-x} \) can be written as \( \frac{1}{t} \):
\[ t + \frac{1}{t} = \frac{10}{3} \]

### Step 4: Multiply through by \( t \)
To eliminate the fraction, multiply through by \( t \):
\[ t^2 + 1 = \frac{10}{3}t \]
Rearranging gives us:
\[ t^2 - \frac{10}{3}t + 1 = 0 \]

### Step 5: Clear the fraction
To make calculations easier, multiply the entire equation by 3:
\[ 3t^2 - 10t + 3 = 0 \]

### Step 6: Use the quadratic formula
Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = -10, c = 3 \):
\[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
Calculating the discriminant:
\[ t = \frac{10 \pm \sqrt{100 - 36}}{6} \]
\[ t = \frac{10 \pm \sqrt{64}}{6} \]
\[ t = \frac{10 \pm 8}{6} \]

### Step 7: Solve for \( t \)
This gives us two possible solutions for \( t \):
1. \( t = \frac{18}{6} = 3 \)
2. \( t = \frac{2}{6} = \frac{1}{3} \)

### Step 8: Substitute back for \( x \)
Recall that \( t = 3^x \):
1. For \( t = 3 \):
\[ 3^x = 3 \implies x = 1 \]
2. For \( t = \frac{1}{3} \):
\[ 3^x = \frac{1}{3} \implies x = -1 \]

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