Given

$\rm 7^x + \frac{1}{7^x} = 49 \frac{1}{49}$

Let us consider $\rm a = 7^x$. Substituting the values of a in the above equations gives us

$\rm a + \frac{1}{a} = 49 \frac{1}{49}$

$\rm or, a + \frac{1}{a} = \frac{49 \cdot 49 + 1}{49}$

$\rm or, a + \frac{1}{a} = \frac{2402}{49}$

Multiplying both sides of the equation by a, we get,

$\rm or, a \cdot a + \frac{1}{a} \cdot a = \frac{2402}{49} \cdot a$

$\rm or, a^2 + 1 = \frac{2402a}{49}$

Multiplying both sides of the equation by 49, we get,

$\rm or, a^2 \cdot 49 + 1 \cdot 49 = \frac{2402a}{49} \cdot 49$

$\rm or, 49 a^2 + 49 = 2402a$

Subtracting 2402a from both sides of the equation, we get,

$\rm or, 49a^2 + 49 - 2402a = 2402a - 2402a$

$\rm or, 49a^2 - 2402a + 49 = 0$

The above equation is quadratic in a. We use the factorization method to solve it.

$\rm or, 49a^2 - (2401 + 1)a + 49 = 0$

$\rm or, 49a^2 - 2401a - a + 49 = 0$

$\rm or, 49a (a - 49) - 1 (a - 49) = 0$

$\rm or, (49a - 1)(a -49) = 0$

Either

$\rm 49a - 1 = 0$

$\rm or, 49a = 1$

$\rm or, a = \frac{1}{49}$

Substituting the value, $\rm a = 7^x$, we get,

$\rm or, 7^x = \frac{1}{7^2}$

$\rm or, 7^x = 7^{-2}$

$\rm \therefore x = -2$

Or

$\rm a - 49 = 0$

$\rm or, a = 49$

Substituting the value, $\rm a = 7^x$, we get,

$\rm or, 7^x = 7^2$

$\rm \therefore x = 2$

Hence, the required values of x = {-2, 2}.