Ensino Médio ⇒ Sistema com logaritmos Tópico resolvido

[Felipe22]

{y. x^log_yx = x^5/2
{log₄y . log_y(y-3x) = 1

na eq. 1 cheguei a x = y^2 e x = y^1/2
na eq. 2 cheguei a y = 3x + 4
mas não consegui desenvolver para chegar ao gabarito

gab: (4 , 16)

[petras]

Felipe22,

[tex3]
\mathsf{

log_ 4y . log_ y(y-3x) = 1 \implies \frac{log_y(y)}{log_y4}.log_y(y-3x) = 1 \implies \frac{1}{log_y4}.log_y(y-3x)=1\\
\frac{1}{log_y4}.log_y(y-3x)=\frac{1}{log_y(y-3x)}\therefore y -3x = 4(I)\\
log_y(y.x^{log_yx})=log_yx^{\frac{6}{2}} \implies 1+ log_yx^{log_yx} = \frac{5}{2}log_yx\\
1+ log_yx.log_yx-\frac{5x}{2}=0 (log_yx=k)\\
1+k^2-\frac{5k}{2}=0 \therefore k = 2 \vee k = \frac{1}{2}\\
log_yx=2 \implies x=y^2 \vee log_yx=\sqrt y\\
Em(I):(x=y^2): y - 3y^2 = 4 \implies \cancel{\Delta < 0} \\
(x = \sqrt y): y-3\sqrt y = 4 \implies (y-4)^2=(3\sqrt y)^2 \implies y^2-8y+16 - 9y = 0 \\
y^2-17y+16=0 \therefore \cancel{y = 1} (0<0 \neq 1)\vee \boxed{y = 16}\\
Em(I): 16 - 3x = 4 \implies \boxed{x = 4}\\


















}





[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\mathsf{

log_ 4y . log_ y(y-3x) = 1 \implies \frac{log_y(y)}{log_y4}.log_y(y-3x) = 1 \implies \frac{1}{log_y4}.log_y(y-3x)=1\\
\frac{1}{log_y4}.log_y(y-3x)=\frac{1}{log_y(y-3x)}\therefore y -3x = 4(I)\\
log_y(y.x^{log_yx})=log_yx^{\frac{6}{2}} \implies 1+ log_yx^{log_yx} = \frac{5}{2}log_yx\\
1+ log_yx.log_yx-\frac{5x}{2}=0 (log_yx=k)\\
1+k^2-\frac{5k}{2}=0 \therefore k = 2 \vee k = \frac{1}{2}\\
log_yx=2 \implies x=y^2 \vee log_yx=\sqrt y\\
Em(I):(x=y^2): y - 3y^2 = 4 \implies \cancel{\Delta < 0} \\
(x = \sqrt y): y-3\sqrt y = 4 \implies (y-4)^2=(3\sqrt y)^2 \implies y^2-8y+16 - 9y = 0 \\
y^2-17y+16=0 \therefore \cancel{y = 1} (0<0 \neq 1)\vee \boxed{y = 16}\\
Em(I): 16 - 3x = 4 \implies \boxed{x = 4}\\


















}





[/tex3]

[Felipe22]

Agradeço mais uma vez.