Question

$log _{2} 3+log _{2} (x+1)=log _{2} 6 log_{3} 2+log_{3} (x+1)=1$
Me ajuda pessoal??

Answer 1

Uma:
$log_23+log_2(x+1)=log_26\\ log_2(x+1)=log_26 - log_23\\ log_2(x+1)=log_2 {(\frac{6}{3})}\\ log_2(x+1)=log_2 {2}\\ log_2(x+1)=1\\ (x+1)=2^1\\ x+1=2\\ x=2-1\\ x=1$

Outra:
$log_32+log_3(x+1)=1\\ log_3(2)(x+1)=1\\ log_3(2x+2)=1\\ 2x+2=3^1\\ 2x+2=3\\ 2x=3-2\\ 2x=1\\ x=\frac{1}{2}$