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[tex] 3^{x} + 3^{-x} = \frac{10}{3} [/tex]

Răspuns :

[tex]3^x+3^-^x= \frac{10}{3} [/tex]


[tex]3^x+ \frac{1}{3^x}= \frac{10}{3} [/tex]

Notăm [tex]3^x=t[/tex]

[tex]t+ \frac{1}{t}= \frac{10}{3} [/tex]

aducem la acelasi numitor cu 3t

[tex]3t^2+3=10t 3t^2+3-10t=0 3t^2-10t+3=0 a=3 b=-10 c=3 delta=b^2-4ac = 100-36=64 x1=3 x2=1/3[/tex]

Revenim la substitutie

[tex]3^x=3 x=log_{3} 3=1 3^x= \frac{1}{3} x=log_{3} \frac{1}{3} =0-1=-1[/tex]