Respuesta :
Solution:
Given:
[tex]logx+log(x-2)-log(x+4)=0[/tex]Applying the laws of logarithm,
[tex]\begin{gathered} logM+logN=logMN \\ logM-logN=log(\frac{M}{N}) \\ If\text{ }log_ab=x,\text{ then }a^x=b \end{gathered}[/tex]Hence, the equation becomes;
[tex]\begin{gathered} logx+log(x-2)-log(x+4)=0 \\ \\ Applying\text{ the laws of logarithm,} \\ \\ logM+logN=logMN \\ logM-logN=log(\frac{M}{N}) \\ \\ Then, \\ \\ log\frac{(x(x-2))}{x+4}=0 \\ log\frac{(x^2-2x)}{x+4}=0 \\ \\ Also,\text{ applying the law of logarithm} \\ If\text{ }log_ab=x,\text{ then }a^x=b. \\ \\ Hence, \\ log_{10}\frac{(x^{2}-2x)}{x+4}=0 \\ 10^0=\frac{x^2-2x}{x+4} \\ 1=\frac{x^2-2x}{x+4} \\ Cross\text{ multiplying,} \\ x^2-2x=x+4 \\ \\ Collecting\text{ all the terms to the left-hand side to form a quadratic equation;} \\ x^2-2x-x-4=0 \\ x^2-3x-4=0 \end{gathered}[/tex]Solving the quadratic equation,
[tex]\begin{gathered} x^2-3x-4=0 \\ x^2+x-4x-4=0 \\ x(x+1)-4(x+1)=0 \\ (x-4)(x+1)=0 \\ x-4=0,\text{ }x+1=0\text{ } \\ x=0+4,\text{ }x=0-1 \\ x=4,\text{ }x=-1 \end{gathered}[/tex]Therefore, the correct answer is x = 4
