Answer:
Option C. x = -4
Step-by-step explanation:
We have to find the extraneous solution of the given logarithmic equation.
[tex]log_{3}(18x^{3})-log_{3}(2x)=log_{3}(144)}[/tex]
[tex]log_{3}(\frac{18x^{3}}{2x})=log_{3}(144)[/tex] [Since log ([tex]\frac{a}{b}[/tex])=log a - log b ]
Now [tex]\frac{18x^{3} }{2x}=144[/tex]
[tex]9x^{2} =144[/tex]⇒[tex]x^{2} =\frac{144}{9}=16[/tex]
x = ±4
Now we put x = 4 in the logarithmic equation
[tex]log_{3}(18\times 4^{3})-log_{3}(2\times 4)=log_{3}(144)[/tex]
[tex]log_{3}(\frac{18\times 64}{8})=log_{3}(144)[/tex]
[tex]log_{3} 144=log_{3}144[/tex] So x = 4 is the real solution
Now we put x = -4 in the logarithmic equation
[tex]log_{3}[18(-4)^{3}]-log_{3}[2(-4)]=log_{3}(144)[/tex]
[tex]log_{3}(-1152)-log_{3}(-8)=log_{3}(144)[/tex]
Since logarithm of any negative number is not defined so x = -4 will be the extraneous solution.
