Question

Which equation can you solve to find the potential solutions to the equation log2x + log2(x – 6) = 4?

Answer 1

$\text{The domain:}\\x \ \textgreater \ 0\ \wedge\ x-6 \ \textgreater \ 0\\x \ \textgreater \ 0\ \wedge\ x \ \textgreater \ 6\\\\D:x\in(6;\ \infty)\\\\\log_2x+\log_2(x-6)=4\\\\\log_2[x(x-6)]=\log_22^4\iff x^2-6x=16\\\\x^2-2\cdot x\cdot3=16\ \ \ \ |+3^2\\\\\underbrace{x^2-2\cdot x\cdot3+3^2}_{(a-b)^2=a^2-2ab+b^2}=16+3^2\\\\(x-3)^2=25\to x+3=\pm\sqrt{25}\\\\x-3=-5\ \vee\ x-3=5\ \ \ \ |+3\\\\x=-2\notin D\ \vee\ x=8\in D\\\\Answer:\ x=8$

$\text{Used:}\\\\\log_ab+\log_ac=\log_a(b\cdot c)\\\\\log_ab=c\iff a^c=b$

Answer 2

Answer:

The answer is C on engenuity

Step-by-step explanation:

x2 – 6x – 16 = 0