2^x*3^x=1296

Solution for 2^x*3^x=1296 equation:

2^x*3^x=1296

We move all terms to the left:

2^x*3^x-(1296)=0

Wy multiply elements

6x^2-1296=0

a = 6; b = 0; c = -1296;
Δ = b2-4ac
Δ = 02-4·6·(-1296)
Δ = 31104
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:

$\sqrt{\Delta}=\sqrt{31104}=\sqrt{5184*6}=\sqrt{5184}*\sqrt{6}=72\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72\sqrt{6}}{2*6}=\frac{0-72\sqrt{6}}{12}
=-\frac{72\sqrt{6}}{12}
=-6\sqrt{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72\sqrt{6}}{2*6}=\frac{0+72\sqrt{6}}{12}
=\frac{72\sqrt{6}}{12}
=6\sqrt{6}
$