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:

Here is the website link to the notes of Indices.

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