ANSWER
[tex]C.\text{ }x=1\text{ and }x=-4[/tex]
EXPLANATION
We want to find the solution set to the equation given:
[tex]\log_4(x+3)+\log_4x=1[/tex]
According to the laws of logarithm, we have that:
[tex]\begin{gathered} \log_ax+\log_ay=\log_a(x*y) \\ \\ and \\ \\ \log_aa=1 \end{gathered}[/tex]
Therefore, we can rewrite the equation as follows:
[tex]\log_4[(x+3)*x]=\log_44[/tex]
Since the logarithms on both sides have the same base, it implies that:
[tex](x+3)*x=4[/tex]
Simplify and solve for x:
[tex]\begin{gathered} x^2+3x=4 \\ \\ x^2+3x-4=0 \\ \\ x^2+4x-x-4=0 \\ \\ x(x+4)-1(x+4)=0 \\ \\ (x-1)(x+4)=0 \\ \\ x=1\text{ and }x=-4 \end{gathered}[/tex]
Hence, the correct answer is option C.
