Question: Solve: $3^{x+2} + \dfrac{1}{3^{x-2}} = 30$

Solution:
Given,

$3^{x+2} + \dfrac{1}{3^{x-2}} = 30$

$or, 3^x × 3^2 + \dfrac{1}{3^x × 3^{-2}} = 30$

$or, 3^x × 9 + \dfrac{3^2}{3^x} = 30$

$or, \dfrac{3^x (3^x × 9) + 3^2}{3^x} = 30$

$or, 9× (3^x)^2  + 9  = 30×3^x$

[ Assume 3^x = a ]

$or, 9a² + 9 = 30a$

$or, 9a² -30a +9 = 0$

$or, 9a² -(27+3)a + 9 = 0$

$or, 9a² -27a -3a +9 = 0$

$or, 9a (a -3) -3(a -3) = 0$

$or, (9a -3)(a-3) = 0$

Either,

$9a -3 = 0$

$or, 9a = 3$

$or, a = \dfrac{3}{9}$

$or, a = \dfrac{1}{3}$

$or, 3^x = 3^{-1}$

$\therefore x = -1$

Or,

$a -3 = 0$

$or, 3^x = 3^1$

$\therefore x = 1$

Hence, the possible values of x are 1 and -1.


Related Notes and Solutions:

Here is the website link to the notes of Indices.

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