1.1.4 \( 3^{x+1}+3^{x-1}+3^{x}=39 \)

To tackle the equation \( 3^{x+1} + 3^{x-1} + 3^{x} = 39 \), we can simplify it by factoring out \( 3^{x-1} \). This gives us: \[ 3^{x+1} = 3 \cdot 3^{x} \quad \text{and} \quad 3^{x-1} = \frac{3^{x}}{3} \] Rewriting the equation, we have: \[ 3 \cdot 3^{x} + \frac{3^{x}}{3} + 3^{x} = 39 \] Now, if we denote \( 3^{x} \) as \( y \), the equation transforms into: \[ 3y + \frac{y}{3} + y = 39 \] Multiplying the entire equation by 3 to eliminate the fraction: \[ 9y + y + 3y = 117 \] This simplifies to: \[ 13y = 117 \] Thus, \( y = 9 \), which means \( 3^x = 9 \). Therefore, we find \( x = 2 \). So, the solution is \( x = 2 \).