Question

2 to the power x + 1 + 2 to the power 2 X is equals to 2 ^ 3​

Answer 1

Answer:

Required value of x is 1.

Step-by-step explanation:

Given,

2^( x + 1 ) + 2^( 2x ) = 2^3

= > 2^( x + 1 ) + { 2^( x ) }^2 = 2^3

= > 2^x . 2^1 + ( 2^x )^2 = 2^3

Let, 2^x = a

= > a.2^1 + ( a )^2 = 2^3

= > a.2 + a^2 = 8

= > 2a + a^2 = 8

= > a^2 + 2a - 8 = 0

= > a^2 + ( 4 - 2 )a - 8 = 0

= > a^2 + 4a - 2a - 8 = 0

= > a( a + 4 ) - 2( a + 4 ) = 0

= > ( a + 4 )( a - 2 ) = 0

= > a = - 4 or 2

Here, a can't be negative, a is 2.

= > 2^x = 2

= > 2^x = 2^1

= > x = 1

Hence the required value of x is 1.

Answer 2

Solution :-

$\sf {2}^{x + 1} + {2}^{2x} = {2}^{3} \\ \\ \\ \sf \implies {2}^{x + 1} + {2}^{x(2)} = {2}^{3} \\ \\ \\ \sf \implies {2}^{x + 1} + ( {2}^{x} )^2 = {2}^{3} \qquad \bigg[ \because {a}^{mn} = (a^m )^n \bigg] \\ \\ \\ \sf \implies {2}^{x}({2}^{1} ) + ( {2}^{x})^2 = {2}^{3} \qquad \bigg[ \because {a}^{m + n} = {a}^{m} \times {a}^{n} \bigg]$

$\rm let \ {2}^{x} = y \\ \\ \\ \sf \implies y( {2}^{1}) + y^{2} = {2}^{3} \\ \\ \\ \sf \implies y(2) + y^{2} = 8 \qquad \bigg[ \because {2}^{3} = 8 \bigg] \\ \\ \\ \sf \implies 2y + {y}^{2} = 8 \\ \\ \\ \sf \implies 2y + {y}^{2} - 8 = 0 \\ \\ \\ \tt \underline{rearranging \ the \ terms} \\ \\ \\ \sf \implies {y}^{2} + 2y - 8 = 0 \\ \\ \\ \tt \underline{splitting \ the \ middle \ term} \\ \\ \\ \sf \implies {y}^{2} + 4y - 2y - 8 = 0 \\ \\ \\ \sf \implies y(y + 4) - 2(y + 4) = 0 \\ \\ \\ \sf \implies (y + 4)(y - 2) = 0 \\ \\ \\ \sf \implies y + 4 = 0 \ or \ y - 2 = 0 \\ \\ \\ \sf \implies y = - 4 \ or \ y = 2$

We know that 2^x = y

If y = 2

2^x = 2

⇒ 2^x = 2¹ [Since 2 can be written ad 2¹]

⇒ x = 1. [Since a^m = a^n then m = n]

If y = - 4

2^x = - 4

- 4 cannot be written to 2 to any power so - 4 is not possible here

So the value of x is 1.