Appropriate Question :- Simplify

[tex]\sf \: {3}^{x} + {3}^{x + 1} + {3}^{x + 2} = 13 \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Given equation is

[tex]\sf \: {3}^{x} + {3}^{x + 1} + {3}^{x + 2} = 13 \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \: {a}^{x + y} = {a}^{x} \times {a}^{y} \: }} \\ \\ [/tex]

So, using this identity, we get

[tex]\sf \: {3}^{x} + {3}^{x} \times {3}^{1} + {3}^{x} \times {3}^{2} = 13 \\ \\ [/tex]

[tex]\sf \: {3}^{x}(1 + {3}^{1} + {3}^{2}) = 13 \\ \\ [/tex]

[tex]\sf \: {3}^{x}(1 + 4 + 9) = 13 \\ \\ [/tex]

[tex]\sf \: {3}^{x}(13) = 13 \\ \\ [/tex]

[tex]\sf \: {3}^{x} = \dfrac{13}{13} \\ \\ [/tex]

[tex]\sf \: {3}^{x} =1 \\ \\ [/tex]

[tex]\sf \: {3}^{x} = {3}^{0} \\ \\ [/tex]

[tex]\bf\implies \:x = 0 \\ \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]