2^x*3^x=1728

Solution for 2^x*3^x=1728 equation:

2^x*3^x=1728

We move all terms to the left:

2^x*3^x-(1728)=0

Wy multiply elements

6x^2-1728=0

a = 6; b = 0; c = -1728;
Δ = b2-4ac
Δ = 02-4·6·(-1728)
Δ = 41472
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:

$\sqrt{\Delta}=\sqrt{41472}=\sqrt{20736*2}=\sqrt{20736}*\sqrt{2}=144\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-144\sqrt{2}}{2*6}=\frac{0-144\sqrt{2}}{12}
=-\frac{144\sqrt{2}}{12}
=-12\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+144\sqrt{2}}{2*6}=\frac{0+144\sqrt{2}}{12}
=\frac{144\sqrt{2}}{12}
=12\sqrt{2}
$