Question

log (3x-1) na base raiz de 2 + log x na base raiz de dois = 2

Answer 1

.
$\log_ \sqrt{2}}(3x-1)+\log_{ \sqrt{2} }{x}=2 \\ \\ \log_{ \sqrt{2} }{(3x-1)(x)}=2 \\ \\ x(3x-1)=( \sqrt{2})^2 \\ \\ 3x^2-x=2 \\ \\ 3x^2-x-2=0$

Δ=b²-4ac
Δ=(-1)^2-4(3)(-2)
Δ=1+24
Δ=25

$x= \frac{1\pm \sqrt{25} }{2.3} = \frac{1\pm5}{6} \\ \\ x'= \frac{1+5}{6} = \frac{6}{6} =1 \\ \\ x"= \frac{1-5}{6} =- \frac{4}{6} =- \frac{2}{3} ~~n/~~serve$

condiçao
x>0
3x-1 >0

S={1}