49^(2x)=1/7

Solution for 49^(2x)=1/7 equation:

49^(2x)=1/7

We move all terms to the left:

49^(2x)-(1/7)=0

We add all the numbers together, and all the variables

49^2x-(+1/7)=0

We get rid of parentheses

49^2x-1/7=0

We multiply all the terms by the denominator


49^2x*7-1=0

Wy multiply elements

343x^2-1=0

a = 343; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·343·(-1)
Δ = 1372
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1372}=\sqrt{196*7}=\sqrt{196}*\sqrt{7}=14\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{7}}{2*343}=\frac{0-14\sqrt{7}}{686}
=-\frac{14\sqrt{7}}{686}
=-\frac{\sqrt{7}}{49}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{7}}{2*343}=\frac{0+14\sqrt{7}}{686}
=\frac{14\sqrt{7}}{686}
=\frac{\sqrt{7}}{49}
$