3^x*4^x=1

Solution for 3^x*4^x=1 equation:

3^x*4^x=1

We move all terms to the left:

3^x*4^x-(1)=0

Wy multiply elements

12x^2-1=0

a = 12; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·12·(-1)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*12}=\frac{0-4\sqrt{3}}{24}
=-\frac{4\sqrt{3}}{24}
=-\frac{\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*12}=\frac{0+4\sqrt{3}}{24}
=\frac{4\sqrt{3}}{24}
=\frac{\sqrt{3}}{6}
$