"The powers of 3" - 2

$\begin{aligned} 3^{x+1} - 3^{x-1} & = \frac 89 \\ 3^{x-1}\left(3^2 - 1\right) & = 8 \times 3^{-2} \\ 8 \times 3^{\color{#3D99F6}x-1} & = 8 \times 3^{\color{#3D99F6}-2} \\ \implies \color{#3D99F6} x - 1& = \color{#3D99F6} -2 \\ x & = \boxed{-1} \end{aligned}$

$\begin{aligned} 3^{x+1} - 3^{x-1} &= \frac 89 \\ 9 \cdot 3^{x+1} - 9 \cdot 3^{x-1} &= 8 \\ 9 \cdot {\color{#D61F06}3^{x+1}} - 1 \cdot {\color{#D61F06}3^{x+1}} &= 8 \\ 9{\color{#D61F06}u} - {\color{#D61F06}u} &= 8 \\ {\color{#D61F06}u} &= 1 \\ {\color{#D61F06}3^{x+1}} &= 1 \\ x+1 &= 0 \\ x &= \boxed{-1} \end{aligned}$

LHS= 3^(x+1)-3^(x-1) 3^(x)×3-3^(x)×3^(-1) 3^(x)[3- $frac {1}{3}$ ] 3^(x) $frac {8}{3}$ 3^(x) × $frac {8}{9}$ × 3 3^(x+1) × $frac {8}{9}$

RHS= $frac {8}{9}$

Hence by comparison we get 3^(x+1)=1 3^(x+1)=3^0 Therefore, x+1=0 Hence, x= -1

Note that $\frac{8}{9} = \color{#20A900}1 \color{#333333}- \color{#3D99F6}\frac{1}{9}$

$= \color{#20A900}3^0 \color{#333333}- \color{#3D99F6}3^{-2}$

which gives $x = -1$