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2 la 2x - 13 ori 6 la x-1 +9 la x =0
solutiile sunt ?


Răspuns :

[tex]2^{2x}-13\cdot 6^{x-1}+9^x=0\\ 2^{2x}-13 \cdot 3^{x}\cdot2^{x}\cdot\frac{1}{6}+3^{2x}=0 |\cdot6\\ 6\cdot (2^x)^2-12\cdot 3^x\cdot2^x+6\cdot (3^x)^2-3^x\cdot 2^x=0\\ 6(2^x-3^x)^2=3^x\cdot2^x\\ (2^x-3^x)^2=\frac{6^x}{6}\\ (2^x-3^x)^2=6^{x-1}\\ Nu\ stiu\ mai\ departe\ [/tex]