Question: Solve: 2^x + \dfrac{1}{2^x} = 4\dfrac{1}{4}

Solution:
Given,

2^x + \dfrac{1}{2^x} = 4\dfrac{1}{4}

or, \dfrac{2^x × 2^x +1}{2^x } = \dfrac{4×4 +1}{4}

or, \dfrac{(2^x)^2 +1}{2^x} = \dfrac{17}{4}

or, 4 \{ (2^x)^2 +1) \} = 17 ×2^x

or, 4×(2^x)^2 + 4 = 17 × 2^x

[ Assume 2^x = a ]

or, 4a² +4 = 17a

or, 4a² - 17a +4 = 0

or, 4a² - (16+1) +4 = 0

or, 4a² -16a -a +4= 0

or, 4a(a -4) -1(a-4) = 0

or, (4a -1)(a-4) = 0

Either,

4a -1 = 0

or, 4a = 1

or, a = \dfrac{1}{4}

or, 2^x = \dfrac{1}{2^2}

or, 2^x = 2^{-2}

\therefore x = -2

Or,

a -4 = 0

or, a = 4

or, 2^x = 2^2

\therefore x = 2

Hence, the value of x is either 2 or -2.


Related Notes and Solutions:

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