Question

Please help me solve this question ....ASAP​

Answer 1

✿$\underline{\textbf{\textsf{ \purple{Solution}:- }}}$

$\sf{ {2}^{(x -2)} + {2}^{(3 - x)} = 3}$

$\longrightarrow \sf{ \dfrac{ 2^x }{ 2^2 } + \dfrac{ 2^3 }{ 2^x } = 3}$

Let $\sf{ 2^x }$ be $\sf{ a }$

$\longrightarrow \sf{ \dfrac{a}{4} + \dfrac{8}{a} = 3}$

$\longrightarrow \sf{ \dfrac{ {a}^{2} + 32 }{4a} = 3}$

$\longrightarrow \sf{ a^2 + 32 = 12a }$

$\longrightarrow \sf{ a^2 - 12a + 32 = 0 }$

$\longrightarrow \sf{ a^2 - 4a - 8a + 32 = 0 }$

$\longrightarrow \sf{ a(a - 4) - 8(a - 4) = 0 }$

$\longrightarrow \sf{ (a - 4)(a - 8) = 0}$

$\sf{\\}$

So,

Either,

$\sf{ (a - 4) = 0 }$

$\longrightarrow \underline{\sf{ \blue{a = 4} }}$

$\sf{\\}$

Or,

$\sf{ (a - 8) = 0 }$

$\longrightarrow \underline{\sf{ \blue{a = 8} }}$

$\sf{\\}$

$\underline{\textsf{ \blue{If a = 4:-} }}$

$\sf{ 2^x = a }$

$\longrightarrow \sf{ 2^x = 4 }$

$\longrightarrow \sf {2^x = 2^2 }$

$\longrightarrow \underline{\sf{\green{x = 2 } }}$

$\sf{\\}$

$\underline{\textsf{ \blue{If a = 8:-} }}$

$\sf{ 2^x = a }$

$\longrightarrow \sf{ 2^x = 8 }$

$\longrightarrow \sf {2^x = 2^3 }$

$\longrightarrow \underline{\sf{ \green{x = 3} }}$

$\sf{\\}$

Hence,

The answer is,

$\longrightarrow \underline{\underline{\sf{ \green{x = 2\:or\:x = 3} }}}$