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2^x*4^x=8
We move all terms to the left:
2^x*4^x-(8)=0
Wy multiply elements
8x^2-8=0
a = 8; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·8·(-8)
Δ = 256
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*8}=\frac{-16}{16} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*8}=\frac{16}{16} =1 $
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