Question

a) log3(3-x)=3
b)logx(x+2)=2
c)log2x(3+xlaputerea2)=2

Answer 1

a) DVA:3-x>0
    3-x=27
       x=-24
b)DVA:x>0
    x+2=x^2
    x^2-x-2=0
    x=-1∉DVA
    x=2
c)DVA:x>0
    3+x=4x^2
4x^2-x-3=0
x=-3/4∉DVA
x=1

Answer 2

$\displaystyle a).log_3(3-x)=3 \\ log_3(3-x)=log_33^3 \\ 3-x=3^3 \\ 3-x=27 \\ -x=27-3 \\ -x=24 \\ \boxed{x=-24} \\ b).log_x(x+2)=2 \\ log_x(x+2)=log_xx^2 \\ x+2=x^2 \\ x^2=x+2 \\ x^2-x-2=0 \\ (x+1)(x-2)=0 \\ x+1=0 \Rightarrow x=-1 \\ x-2=0 \Rightarrow \boxed{x=2}$
$\displaystyle c).log_{2x}(3+x^2)=2 \\ log_{2x}(3+x ^2)=log_{2x}(2x)^2 \\ 3+x^2=(2x)^2 \\ 3+x^2=4x^2 \\ 3+x^2-4x^2=0 \\ -3x^2+3=0| \cdot (-1) \\ 3x^2-3=0 \\ \Delta=0-4 \cdot 3 \cdot (-3)=0+36=36\ \textgreater \ 0 \\ x_1= \frac{0+ \sqrt{36} }{6} = \frac{6}{6} =1 \\ x_2= \frac{0- \sqrt{36} }{6} =- \frac{6}{6} =-1 \\ \boxed{x_1=1}$