Respuesta :

Answer:

B

Step-by-step explanation:

Look at the expression inside the log( ).  Work from the outside in, so to speak.  First, there's a quotient (the cube root divided by 3x).

[tex]\log \left( \frac{\sqrt[3]{2-x}}{3x} \right) = \log\left(\sqrt[3]{2-x}}\right) - \log{3x}[/tex]

The cube root can be written using a rational exponent.

[tex]\log\left(\sqrt[3]{2-x}}\right) - \log{3x} = \log \left( (2-x)^{1/3} \right) - \log{3x}[/tex]

Use the exponent property to move the exponent 1/3 out.

[tex]\log (2-x)^{1/3} - \log{3x} = \frac{1}{3} \log{(2-x)} - \log{3x}[/tex]

This now matches answer choice B.

ACCESS MORE
EDU ACCESS