Question

Please I really need help with this

Answer 1

Answer:

Step-by-step explanation:

㏒ 3+㏒(x+2)=1

㏒3(x+2)=1

3(x+2)=10^1

3x+6=10

3x=10-6=4

x=4/3

so A

3.

$log_{2}x+log_{2}(x+2)=3\\or~log_{2}[x(x+2)]=3\\x(x+2)=2^3\\x^2+2x-8=0\\x^2+4x-2x-8=0\\x(x+4)-2(x+4)=0\\(x+4)(x-2)=0\\x=2,-4\\x=-4 ~is~an~extraneous~solution.\\B$