3^x*2^x=16

Solution for 3^x*2^x=16 equation:

3^x*2^x=16

We move all terms to the left:

3^x*2^x-(16)=0

Wy multiply elements

6x^2-16=0

a = 6; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·6·(-16)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*6}=\frac{0-8\sqrt{6}}{12}
=-\frac{8\sqrt{6}}{12}
=-\frac{2\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*6}=\frac{0+8\sqrt{6}}{12}
=\frac{8\sqrt{6}}{12}
=\frac{2\sqrt{6}}{3}
$