Question

HELP!!! Please solve this logarithmic equation for the value of the equation. Be sure to check for extraneous solutions. Thank you!

Answer 1

`D:2x > 0\wedge x+2 > 0\\D:x > 0 \wedge x > -2\\D: x > 0`

`\log_42x+\log_4(x+2)=2\\\log_4(2x(x+2))=2\\2x^2+4x=4^2\\2x^2+4x-16=0\\x^2+2x-8=0\\x^2+2x+1-9=0\\(x+1)^2=9\\x+1=3 \vee x+1=-3\\x=2 \vee x=-4`

`-4\\ot\in D`

Therefore
`x=2`.