Solution
We will solve the question using the laws of logarithm
Some of the laws we will use are stated below
[tex]\log_xa+\log_xb=\log_x(a\times b)[/tex]
Also ,
[tex]\begin{gathered} \log_xy=z \\ \\ x^z=y \end{gathered}[/tex]
Given ;
[tex]\log_4(x-3)+\log_4(x+3)=2[/tex]
Using same logarithm base ruke , we have
[tex]\log_4(x-3)(x+3)=2[/tex]
Removing the logarithm , we have
[tex]4^2=(x-3)(x+3)[/tex]
We can now solve the equation obtained as shown below
[tex]\begin{gathered} 16=x^2+3x-3x-9 \\ 16=x^2-9 \\ x^2=16+9 \\ x^2=25 \\ x=\pm\sqrt{25} \\ x=\pm5 \end{gathered}[/tex]
We will ignore x is negative 5 , since we cannot have a negative logarithm
The final answer is x=5
OPTION B
