Question

AJUUUUDAAAA!!!! OBRIGADO DESDE JÁ

$a) <var>log _{4} x + log _{4} x^{2}= 12</var>$

$b) <var>(log x)^{2} - 3logx= -2</var>$

$c) <var>log _{2}(x-1)^{2} - 5 log (x-1)=0</var>$

Answer 1

$a) \ log_4x+log_4x^2=12\\ \\ log_4(x.x^2)=12\\ \\ log_4x^3=12\\ \\ x^3=4^{12}\\ \\ x= \sqrt[3]{4^{12}}\\ \\ x=4^4\\ \\ \boxed{x= 256}$

$b) \ log^2x-3logx=-2\\ \\ \boxed{y=logx\\} \\ y^2-3y+2=0\\ \\ y_1=1\\ \\ y_2=2\\ \\ logx=1 \rightarrow x=10\\ \\ logx=2\rightarrow x=100$

$c) \ log_2(x-1)^2-5log_2(x-1)=0\\ \\ log_2(x-1)^2=5log_2(x-1)\\ \\ log_2(x-1)^2=log_2(x-1)^5\\ \\ (x-1)^2=(x-1)^5\\ \\ \boxed{x=2}$

Answer 2

a) $log_{4} x+log_{4} x^{2} =12$

$log_{4} x+2.log_{4} x =12$

$3.log_{4} x =12$

$log_{4} x =4$

$x= 4^{4}$

$x=256$


b) $(logx)^{2}-3logx=-2$

$(logx)^{2}-3logx+2=0$

$logx=y$

$y^{2}-3y+2=0$

$(y-1).(y-2)=0$

$y-1=0$  ⇒  $y=1$  ⇒ $logx=1$  ⇒ $x= 10^{1} =10$

ou

$y-2=0$  ⇒ $y=2$  ⇒  $logx=2$  ⇒ $x= 10^{2} =100$


c) $log_{2} (x-1)^{2} -5log_{2} (x-1)=0$

$2log_{2} (x-1)-5log_{2} (x-1)=0$

$-3log_{2} (x-1)=0$

$log_{2} (x-1)=0$

$x-1= 2^{0}$

$x-1= 1$

$x=2$