Question

log (10x^2+5)-2log(1-x)=1

Answer 1

$\log(10x^2+5)-2\log(1-x)=1\\\\\log(10x^2+5)-\log(1-x)^2=\log(10)\\\\\log\left(\dfrac{10x^2+5}{(1-x)^2}\right)=\log(10)\\\\ \dfrac{10x^2+5}{(1-x)^2}=10\\\\10x^2+5=10(1-x)^2\\\\10x^2+5=10(1-2x+x^2)\\\\10x^2+5=10-20x+10x^2\\\\20x=5\\\\x=\dfrac{5}{20}=\boxed{\dfrac14}\\\\\\\log\left(10\left(\dfrac14\right)^2+5\right)-2\log\left(1-\dfrac14\right)=1\\\\\log\left(\dfrac{45}{8}\right)-2\log\left(\dfrac34\right)=1\\\\ 1=1\ (V)$

Answer 2

log (10x²+5)= 2log(1-x) +log 10

log (10x²+5)= log(1-x)² +log 10

log (10x²+5)= log 10*(1-x)²

10x²+5=10*(1-x)²

10x²+5=10-20x+x²

5=10-20x

-5=-20x

x=5/20=1/4

Verificando:  log[a] b   , a e b tem que ser maior que zero

Para x=1/4 ==> 10x²+5>0  ....OK

Para x=1/4 ==>1-x  = 1-1/4 =3/4>0  ...OK

Resposta x=1/4