Answer :
9514 1404 393
Answer:
C. 4/9
Step-by-step explanation:
There are a couple of ways you can do this.
[tex]\log_3{4x}-2\log_3{x}=2\\\\\log_3{4}+\log_3{x}-2\log_3{x}=2\\\\\log_3{4}-2=\log_3{x}\\\\4\cdot3^{-2}=x\qquad\text{take antilogs}\\\\\boxed{x=\dfrac{4}{9}}\\\\\textsf{or}\\\\\dfrac{4x}{x^2}=3^2\qquad\text{take antilogs}\\\\\dfrac{4}{9}=x\qquad\text{cancel $x$, multiply by $\dfrac{x}{9}$}[/tex]