Question

$8 \times {2}^{2x} + 4 \times {2}^{x + 1} = 1 + {2}^{x}$
​

Answer 1

$\large\underline\blue{\bold{Given \:  Question  }}$

$\pink{ \tt \: 8 \times {2}^{2x} + 4 \times {2}^{x + 1} = 1 + {2}^{x}}$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\purple{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\orange{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\color{peru}{(a^m)^n\:=\:a^{m\times{n}}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\red{ {x}^{0} = 1}}}}} \\ \end{gathered}$

$\begin{gathered}(6)\:{\underline{\boxed{\bf{\blue{a^m \: = \: {a^n} \: \:then\: \: {m\: = \:n}\:}}}}} \\ \end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \: :\implies \: 8 \times {2}^{2x} + 4 \times {2}^{x + 1} = 1 + {2}^{x}$

$\tt \: :\implies \: 8 \times {2}^{2x} + 4 \times {2}^{x} \times 2= 1 + {2}^{x}$

$\tt \: :\implies \: 8 \times { ({2}^{x} )}^{2} + 8 \times {2}^{x} = 1 + {2}^{x}$

$\blue{ \: \tt \: :\implies \: Let \: {2}^{x} = y}$

$\tt \: :\implies \: {8y}^{2} + 8y = 1 + y$

$\tt \: :\implies \: {8y}^{2} + 8y - y - 1 = 0$

$\tt \: :\implies \: {8y}^{2} + 7y - 1 = 0$

$\tt \: :\implies \: {8y}^{2} + 8y - y - 1 = 0$

$\tt \: :\implies \: 8y(y + 1) - 1(y + 1) = 0$

$\tt \: :\implies \: (8y - 1)(y + 1) = 0$

$\tt \: :\implies \: 8y - 1 = 0 \: \: or \: \: y + 1 = 0$

$\tt \: :\implies \: y = \dfrac{1}{8} \: \: or \: \: y \: = \: - \: 1$

$\tt \: :\implies \: {2}^{x} = \dfrac{1}{8} \: \: or \: \: {2}^{x} = - 1 \: \pink{ \bf(rejected)}$

$\tt \: :\implies \: {2}^{x} = {2}^{ - 3}$

$\large \boxed{ \pink{ \tt \: :\implies \: x \: = \: - \: 3}}$

Verification :-

Consider LHS,

We have,

$\tt \: :\implies \: 8 \times {2}^{2x} + 4 \times {2}^{x + 1}$

On substituting x = - 3, we get

$\tt \: :\implies \: 8 \times {2}^{2( - 3)} + 4 \times {2}^{ - 3 \: + \: 1}$

$\tt \: :\implies \: 8 \times {(2)}^{ - 6} + 4 \times {(2)}^{ - 2}$

$\tt \: :\implies \: \dfrac{8}{ {2}^{6} } + \dfrac{4}{ {2}^{2} }$

$\tt \: :\implies \: \dfrac{8}{64} + \dfrac{4}{4}$

$\tt \: :\implies \: \dfrac{1}{8} + 1$

$\tt \: :\implies \: \dfrac{9}{8}$

Now,

Consider RHS,

We have,

$\tt \: :\implies \: 1 + {2}^{x}$

On substituting x = - 3, we get

$\tt \: :\implies \: 1 + {(2)}^{ - 3}$

$\tt \: :\implies \: 1 + \dfrac{1}{ {2}^{3} }$

$\tt \: :\implies \: 1 + \dfrac{1}{8}$

$\tt \: :\implies \: \dfrac{9}{8}$

➦ LHS = RHS

$\large{\boxed{\boxed{\bf{Hence, verified}}}}$