Ensino Médio ⇒ equação logaritmica

[Felipe22]

log _{1/[tex3]\sqrt{1+x}[/tex3]} 10.log₁₀(x²-3x +2) = -2 + log _{1/ [tex3]\sqrt{1+x}[/tex3]} 10.log₁₀(x-3)

S:{ 5 }

Tentei várias vezes e infelismente não consegui.

[petras]

Felipe22,
[tex3]\mathsf{
C.E.: x > 3; \\
log_{\frac{1}{\sqrt{1+x}}}(10).log[(x^2-3x+2)]=-2+log_{\frac{1}{\sqrt{1+x}}}(10).log(x-3)\\
\frac{log10}{log(\frac{1}{\sqrt{1+x}})}.log(x^2-3x+2)=-2+\frac{log10}{log(\frac{1}{\sqrt{1+x}})}.log(x-3)\\
\frac{1}{-log(1+x)^\frac{1}{2}}.log(x^2-3x+2) = -2+\frac{1}{-log(1+x)^\frac{1}{2}}.log(x-3)=\\
\frac{-2}{log(1+x)}.log(x^2-3x+2) = -2-\frac{2}{log(1+x)}.log(x-3)=\\

-2log(x^2-3x+2) = -2[log(1+x)+2log(x-3)]\\
-2log(x^2-3x+2) = -2[log(1+x).(x-3)]\\
log(x^2-3x+2) = log(x^2-2x+-3)\\
\therefore
x^2-3x+2=x^2-2x-3 \implies \boxed{x=5}




}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
C.E.: x > 3; \\
log_{\frac{1}{\sqrt{1+x}}}(10).log[(x^2-3x+2)]=-2+log_{\frac{1}{\sqrt{1+x}}}(10).log(x-3)\\
\frac{log10}{log(\frac{1}{\sqrt{1+x}})}.log(x^2-3x+2)=-2+\frac{log10}{log(\frac{1}{\sqrt{1+x}})}.log(x-3)\\
\frac{1}{-log(1+x)^\frac{1}{2}}.log(x^2-3x+2) = -2+\frac{1}{-log(1+x)^\frac{1}{2}}.log(x-3)=\\
\frac{-2}{log(1+x)}.log(x^2-3x+2) = -2-\frac{2}{log(1+x)}.log(x-3)=\\

-2log(x^2-3x+2) = -2[log(1+x)+2log(x-3)]\\
-2log(x^2-3x+2) = -2[log(1+x).(x-3)]\\
log(x^2-3x+2) = log(x^2-2x+-3)\\
\therefore
x^2-3x+2=x^2-2x-3 \implies \boxed{x=5}




}[/tex3]