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.

#SciPiPupil