Math, asked by ritamduttatejhati, 24 days ago

Find the value of x ( don't scam please)​

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Answered by mathdude500
6

Given Question :-

Solve for x :

\rm :\longmapsto\: {2}^{x - 2} +  {2}^{3 - x} = 3

\purple{ \bf{ \: \large\underline{\sf{Solution-}}}}

Given equation is

\rm :\longmapsto\: {2}^{x - 2} +  {2}^{3 - x} = 3

We know,

 \red{\boxed{ \rm{ \:  {a}^{x - y}  =  {a}^{x}  \div  {a}^{y} \:  \: }}}

So, using this, we get

\rm :\longmapsto\:\dfrac{ {2}^{x} }{ {2}^{2} }  + \dfrac{ {2}^{3} }{ {2}^{x} }  = 3

\rm :\longmapsto\:\dfrac{{2}^{x}}{4}  + \dfrac{8}{{2}^{x}}  = 3

Let we assume that,

 \red{\rm :\longmapsto\:\boxed{ \rm{ \: {2}^{x} = y \:  \: }}}

So, above equation can be rewritten as

\rm :\longmapsto\:\dfrac{y}{4}  + \dfrac{8}{y}  = 3

\rm :\longmapsto\:\dfrac{ {y}^{2}  + 32}{4y} = 3

\rm :\longmapsto\: {y}^{2} + 32 = 12y

\rm :\longmapsto\: {y}^{2} - 12y + 32 = 0

\rm :\longmapsto\: {y}^{2} - 8y - 4y+ 32 = 0

\rm :\longmapsto\:y(y - 8) - 4(y - 8) = 0

\rm :\longmapsto\:(y - 8)(y - 4) = 0

\bf\implies \:y = 8 \:  \:  \: or \:  \:  \: y = 4

\bf\implies \:{2}^{x} = 8 \:  \:  \: or \:  \:  \: {2}^{x} = 4

\bf\implies \:{2}^{x} =  {2}^{3}  \:  \:  \: or \:  \:  \: {2}^{x} =  {2}^{2}

 \red{\bf\implies \:\boxed{ \bf{ \: x = 3 \:  \:  \: or \:  \:  \: x = 2 \:  \: }}}

Verification :-

Case :- 1 When x = 3

 \green{\rm :\longmapsto\: {2}^{x - 2} +  {2}^{3 - x} = 3}

On substituting the value of x, we get

 \green{\rm :\longmapsto\: {2}^{3 - 2} +  {2}^{3 - 3} = 3}

 \green{\rm :\longmapsto\: {2}^{1} +  {2}^{0} = 3}

 \green{\rm :\longmapsto\:2 + 1 = 3}

 \green{\rm :\longmapsto\:3 = 3}

Hence, Verified

Case :- 2 When x = 2

 \green{\rm :\longmapsto\: {2}^{x - 2} +  {2}^{3 - x} = 3}

On substituting the value of x, we get

 \green{\rm :\longmapsto\: {2}^{2 - 2} +  {2}^{3 - 2} = 3}

 \green{\rm :\longmapsto\: {2}^{0} +  {2}^{1} = 3}

 \green{\rm :\longmapsto\: 1 + 2= 3}

 \green{\rm :\longmapsto\: 3= 3}

Hence, Verified

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