Question

Pls Help, 50 Points!

Answer 1

x = 0

Explanation:

`\large \boxed{\begin{minipage}{8 cm}\underline{Log Laws} \\ \\$\textsf{Product law}: \quad \log_ax + \log_ay=\log_a(xy) \\ \\\textsf{If }\: \log_ab=c\: \textsf{ then }\: a^c=b$\\\end{minipage}}`

`\large \begin{aligned}\log_9(x+3)+\log_9(x+1) & = (1)/(2)\\\\\log_9[(x+3)(x+1)] & = (1)/(2)\\\\9^{(1)/(2)} & =(x+3)(x+1)\\\\3 & = x^2+4x+3\\\\x^2+4x & = 0\\\\x(x+4) & = 0 \\\\\implies x & = 0, -4\end{aligned}`

Cannot take logs of negative numbers, therefore x = 0 only.