Ensino Médio ⇒ equação logaritmica Tópico resolvido

[Felipe22]

log₃(3^x - 1).log₃(3^x+1 - 3) = 6

gab: log₃10 ; log₃ 28/27

[petras]

Felipe22,

[tex3]C.E:\\
\boxed{3^x-1>0 \implies 3^x>1 \implies 3^x >3^0 \therefore x > 0\\
3.3^x-3 > 0 \implies 3^x>1 \therefore x > 0 }
\\log_3(3^x - 1).log_3(3^{x+1} - 3) = 6\\
log_3(3^x - 1).log_3(3.3^x - 3) = 6\\
log_3(3^x - 1).log_3(3.(3^x - 1) = 6\\\\
3^x-1=a(I)\\
log_3(a).log_3(3a)=6\\
log_3(a).(log_3(3)+log_3a)=6\\
log_3(a).(1+lg_3(a))=6\\
log_3a+log^2_3a=6\\
log_3a = y\\
y^2+y-6=0 \implies y = -3
\vee y = 2\\
log_3a=-3 \implies a=3^{-3} = \frac{1}{27}\\
Em(I) \therefore 3^x-1 =\frac{1}{27} \implies 3^x = \frac{28}{27} \implies log_33^x = log_3\frac{28}{27}\\
\therefore \boxed{x = log_3\frac{28}{27}}\\
log_3a = 2 \implies a = 9\\
Em(I): 3^x-1=a \implies 3^x - 1 = 9\implies 3^x = 10\\
\therefore log_33x = log_310 \implies \boxed{x = log_310}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]C.E:\\
\boxed{3^x-1>0 \implies 3^x>1 \implies 3^x >3^0 \therefore x > 0\\
3.3^x-3 > 0 \implies 3^x>1 \therefore x > 0 }
\\log_3(3^x - 1).log_3(3^{x+1} - 3) = 6\\
log_3(3^x - 1).log_3(3.3^x - 3) = 6\\
log_3(3^x - 1).log_3(3.(3^x - 1) = 6\\\\
3^x-1=a(I)\\
log_3(a).log_3(3a)=6\\
log_3(a).(log_3(3)+log_3a)=6\\
log_3(a).(1+lg_3(a))=6\\
log_3a+log^2_3a=6\\
log_3a = y\\
y^2+y-6=0 \implies y = -3
\vee y = 2\\
log_3a=-3 \implies a=3^{-3} = \frac{1}{27}\\
Em(I) \therefore 3^x-1 =\frac{1}{27} \implies 3^x = \frac{28}{27} \implies log_33^x = log_3\frac{28}{27}\\
\therefore \boxed{x = log_3\frac{28}{27}}\\
log_3a = 2 \implies a = 9\\
Em(I): 3^x-1=a \implies 3^x - 1 = 9\implies 3^x = 10\\
\therefore log_33x = log_310 \implies \boxed{x = log_310}
[/tex3]

[Felipe22]

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