Question: Solve: $2^{a-2} + 2^{3-a} = 3$


Solution:
Given,

$2^{a-2} + 2^{3-a} = 3$

$or, 2^a × 2^{-2} + 2^3 × 2^{-a} = 3$

$or, 2^a × \dfrac{1}{2^2} + 2^3 × \dfrac{1}{2^a} = 3$

$or, \dfrac{2^a × 2^a + 2^3 × 2^2}{2^2 × 2^a} = 3$

$or, 2^{a+a} + 2^{3+2} = 3×2^2 × 2^a$

$or, (2^a)^2 + 32 = 12 × 2^a$

[ Let 2^a = x ]

$or, (x)^2 +32 - 12x = 0$

$or, x^2 - (8+4)x + 32 = 0$

$or, x^2 -8x -4x +32 = 0$

$or, x(x -8) -4(x-8) = 0$

$or, (x-4)(x-8) = 0$

Either,

$(x -4) = 0$

$or, 2^a = 4$

$or, 2^a = 2^2$

$\therefore a = 2$

Or,

$(x-8) = 0$

$or, 2^a = 8$

$or, 2^a = 2^3$

$\therefore a = 3$

Hence, a = 2 or 3

Related Notes and Solutions:

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