6^x+1 +35•6^x-1=71

To solve the equation \( 6^{x+1} + 35 \cdot 6^{x-1} = 71 \), we can start by rewriting the terms for easier manipulation.

First, we can express \( 6^{x+1} \) as \( 6^x \cdot 6 \) and \( 35 \cdot 6^{x-1} \) as \( 35 \cdot \frac{6^x}{6} \):

\[ 6 \cdot 6^x + 35 \cdot \frac{6^x}{6} = 71 \]

Simplifying \( 35 \cdot \frac{6^x}{6} \) gives us:

\[ 6 \cdot 6^x + \frac{35}{6} \cdot 6^x = 71 \]

Now, combine the terms on the left side:

\[ \left(6 + \frac{35}{6}\right) 6^x = 71 \]

To combine \( 6 + \frac{35}{6} \), we first find a common denominator (which is 6):

\[ 6 = \frac{36}{6} \quad \text{so} \quad 6 + \frac{35}{6} = \frac{36 + 35}{6} = \frac{71}{6} \]

Now, substituting this back into the equation, we have:

\[ \frac{71}{6} 6^x = 71 \]

To eliminate 71 from both sides, we can divide by 71:

\[ \frac{6^x}{6} = 1 \]

Multiplying both sides by 6 gives us:

\[ 6^x = 6 \]

This means:

\[ x = 1 \]

Thus, the solution to the equation \( 6^{x+1} + 35 \cdot 6^{x-1} = 71 \) is:

\[ \boxed{1} \]