Question: $5^x + 5^{-x} = 25\dfrac{1}{25}$


Solution:
Given,

$5^x + 5^{-x} = 25\dfrac{1}[25}$

$or, 5^x + \dfrac{1}{5^x} = 25\dfrac{1}{25}$

$or, \dfrac{5^x × 5^x + 1}{5^x} = \dfrac{25×25 +1}{25}$

$or, \dfrac{(5^x)^2 + 1}{5^x} = \dfrac{626}{25}$

$or, 25\{ (5^x)^2 +1\} = 5^x × 626$

[ Let 5^x = a ]

$or, 25 \{ a² + 1\} = 626a$

$or, 25a² + 25 = 626a$

$or, 25a² -626a +25= 0$

$or, 25a² - (625 +1)a +25 = 0$

$or, 25a² -626a -a +25 = 0$

$or, 25a (a -25) - 1(a -25) = 0$

$or, (25a -1)(a-25) = 0$

Either,

$25a -1 = 0$

$or, 25 × 5^x = 1$

$or, 5^x = \dfrac{1}{25}$

$or, 5^x = \dfrac{1}{5^2}$

$or, 5^x = 5^{-2}$

$\therefore x = -2$

Or,

$a -25 = 0$

$or, a = 25$

$or, 5^x = 5^2$

$\therefore x = 2$

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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