6^x+1 +35•6^x-1=71

Let's solve the equation \(6^{x+1} + 35 \cdot 6^{x-1} = 71\).

First, let's rewrite \(6^{x+1}\) and \(6^{x-1}\):

• \(6^{x+1} = 6 \cdot 6^x\)
• \(6^{x-1} = \frac{6^x}{6}\)

Substituting these into the equation gives: \[ 6 \cdot 6^x + 35 \cdot \frac{6^x}{6} = 71 \] This simplifies to: \[ 6 \cdot 6^x + \frac{35}{6} \cdot 6^x = 71 \]

Now combine the terms on the left side: \[ (6 + \frac{35}{6}) \cdot 6^x = 71 \]

To combine the constants, we find a common denominator: \[ 6 = \frac{36}{6} \] So, \[ \frac{36}{6} + \frac{35}{6} = \frac{71}{6} \] Thus, we have: \[ \frac{71}{6} \cdot 6^x = 71 \]

To isolate \(6^x\), divide both sides by \(\frac{71}{6}\): \[ 6^x = 71 \cdot \frac{6}{71} = 6 \]

Now we take the logarithm of both sides or recognize that \(6^1 = 6\): \[ x = 1 \]

Thus, the solution to the equation is: \[ \boxed{1} \]