Question: Solve: $7^x + \dfrac{343}{7^x} = 56$


Solution:
Given,

$7^x + \dfrac{343}{7^x} = 56$

$or, \dfrac{7^x × 7^x + 343}{7^x} = 56$

$or, 7^{x +x} + 343 = 56 × 7^x$

$or, 7^{2x} + 343 = 56 × 7^x$

$or, (7^x)^2 + 343 = 56×7^x$

[ Assume 7^x as a ]

$or, a² + 343 = 56a$

$or, a² -56a +343 = 0$

$or, a² -(49+7)a + 343 = 0$

$or, a² -49a - 7a +343 = 0$

$or, a(a -49) - 7(a -49) = 0$

$or, (a -7)(a -49) = 0$

Either,

$a -7 = 0$

$or, 7^x = 7^1$

$\therefore x=1$

Or,

$a -49 = 0$

$or, 7^x = 49$

$or, 7^x = 7^2$

$\therefore x = 2$

Hence, the values of x are 1 or 2.


Related Notes and Solutions:

Here is the website link to the notes of Indices.

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