Respuesta :
[tex]{ \overline{ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}[/tex]
[tex] \large \underline \mathbb{PROBLEM : }[/tex]
[tex] \qquad\quad\tt{Solve \: for \: x} : \\ \\\qquad\quad \sf \: \bigg( \frac{1}{3}\bigg)^{ - 4} \times \bigg( \frac{1}{3}\bigg)^{ - 8} = \bigg( \frac{1}{3}\bigg)^{ - 2x} \\ \\ \\ [/tex]
[tex]{ \overline{ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}[/tex]
[tex] \large \underline \mathbb{ANSWER : }[/tex]
[tex] \qquad \qquad \quad\qquad \huge \tt{x = 6 }\\ \\ [/tex]
[tex]{ \overline{ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}[/tex]
[tex] \large \underline \mathbb{SOLUTION : }[/tex]
Given :-
[tex]\qquad\quad\large\sf \: \bigg( \frac{1}{3}\bigg)^{ - 4} \times \bigg( \frac{1}{3}\bigg)^{ - 8} = \bigg( \frac{1}{3}\bigg)^{ - 2x} \\ \\ [/tex]
We know :-
[tex] \qquad \qquad\large\boxed{ \sf{ \: {a}^{x} \times {a}^{y} = {a}^{x + y} \: }} \\ \\ [/tex]
We get :-
[tex]\large\sf \Longrightarrow \qquad \bigg( \frac{1}{3}\bigg)^{ - 4 + ( - 8)} = \bigg( \frac{1}{3}\bigg)^{ - 2x} \\ \\ [/tex]
[tex] \large\sf \Longrightarrow \qquad \bigg( \frac{1}{3}\bigg)^{ - 4 - 8} = \bigg( \frac{1}{3}\bigg)^{ - 2x} \\ \\[/tex]
[tex]\large\sf \Longrightarrow \qquad \bigg( \frac{1}{3}\bigg)^{ - 12} = \bigg( \frac{1}{3}\bigg)^{ - 2x} \\ \\ [/tex]
We know :-
[tex]\large \qquad \qquad\boxed{ \sf{ \: {a}^{x} = {a}^{y} \: \sf \: \implies \: x = y \: }} \\ \\[/tex]
So,
[tex]\large\sf\Longrightarrow \quad - 12 = - 2x \\ \\[/tex]
[tex]\large\sf\Longrightarrow \quad \underline{\sf {\red{x = 6}}} \: \sf{ --- \: Final \: Answer} \\ \\[/tex]
[tex]\underline{\rule{190pt}{8pt}} \\[/tex]
ᜎᜒ [tex]\large \mathfrak{TheFletchhhh \: \: X'D}[/tex]
