Easy exponents #7

Just factorize it:

${ 4 }^{ x-1 }+{ 4 }^{ x }+{ 4 }^{ x+1 }=84\\ \\ { 4 }^{ x }\left( \frac { 1 }{ 4 } +1+4 \right) =84\\ \\ { \quad \quad 4 }^{ x }\times \left( \frac { 21 }{ 4 } \right) =84\\ \\ \quad \quad \quad \quad \quad \quad { \quad 4 }^{ x }=16\\ \\ \quad \quad \quad \quad \quad \quad \Rightarrow \boxed { x=2 }$

$\begin{aligned} 4^{x - 1} + 4^x × 4^{x + 1} & = 84\\ 4^x\left(\dfrac{1}{4} + 1 + 4 \right) & = 84\\ 4^x × \dfrac{21}{4} & = 84\\ 4^x & = 16\\ 4^x & = 4^2\\ \therefore x & = \color{#69047E}{\boxed{2}} \end{aligned}$