Question

Select the correct answer.

Answer 1

9514 1404 393

Answer:

  C.  4/9

Step-by-step explanation:

There are a couple of ways you can do this.

  $\log_3{4x}-2\log_3{x}=2\\\\\log_3{4}+\log_3{x}-2\log_3{x}=2\\\\\log_3{4}-2=\log_3{x}\\\\4\cdot3^{-2}=x\qquad\text{take antilogs}\\\\\boxed{x=\dfrac{4}{9}}\\\\\textsf{or}\\\\\dfrac{4x}{x^2}=3^2\qquad\text{take antilogs}\\\\\dfrac{4}{9}=x\qquad\text{cancel x , multiply by \dfrac{x}{9} }$