Question: Solve: $2^x + \dfrac{16}{2^x} = 10$


Solution:
Given,

$2^x + \dfrac{16}{2^x} = 10$

$or, \dfrac{2^x × 2^x +16}{2^x} = 10$

$or, (2^x)^2 + 16 = 10×2^x$

[ Let 2^x = a ]

$or, a² + 16 = 10a$

$or, a² -10a +16 = 0$

$or, a² - (8+2)a  +16 = 0$

$or, a² -8a -2a +16 = 0$

$or, a(a -8) -2(a -8) = 0$

$or, (a-2)(a-8) = 0$

Either,

$a -2 = 0$

$or, 2^x = 2^1$

$\therefore x = 1$

Or,

$a -8 = 0$

$or, 2^x = 8$

$or, 2^x = 2^3$

$\therefore x = 3$

Hence, the value of x is either 1 or 3.


Related Notes and Solutions:

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