Respuesta :

caylus

Answer:

x=2 or x=3

Step-by-step explanation:

[tex]2^{x-2}+2^{3-x}=3\\\\\dfrac{2^x}{4} +\dfrac{8}{2^x}=3 \\\\\dfrac{2^{2x}+8*4}{4*2^x}=3\\ \\(2^x)^2-12*2^x+32=0\\\\\Delta=12^2-4*32=16=4^2\\\\2^x=\dfrac{12-4}{2}=4\ or\ 2^x=\dfrac{12+4}{2}=8\\\\2^x=2^2 \ = > x=2\ or\ 2^x=2^3 \ = > x=3\\[/tex]

Thanks to rishiraj39 to delete my answer to the question .

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