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zoth

16 Ian 2026 14:00



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Am gasit una nostima pt Catalin:

Daca: 2^x+3^x=16 si

(6^x+4^x)/4^x=2, gasiti-l pe x

Solutie propusa:

(6^x+4^x)/4^x=2 

[(2*3)^x+(2*2)^x]/(2*2)^x=2

(2^x*3^x+2^x*2^x)/2^x*2^x=2

2^x(3^x+2^x)/2^x*2^x=2

(2^x+3^x)/2^x=2   dar 2^x+3^x=16

16/2^x=2

16=2*2^x

8=2^x

2^3=2^x

x=3

Solutia x=3 nu verifica nici prima nici a doua ecuatie.
Care e duda?