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2^x*3^x=8
We move all terms to the left:
2^x*3^x-(8)=0
Wy multiply elements
6x^2-8=0
a = 6; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·6·(-8)
Δ = 192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*6}=\frac{0-8\sqrt{3}}{12} =-\frac{8\sqrt{3}}{12} =-\frac{2\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*6}=\frac{0+8\sqrt{3}}{12} =\frac{8\sqrt{3}}{12} =\frac{2\sqrt{3}}{3} $
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