Answer:
B
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).
![\log \left( \frac{\sqrt[3]{2-x}}{3x} \right) = \log\left(\sqrt[3]{2-x}}\right) - \log{3x}](https://img.qammunity.org/2022/formulas/mathematics/high-school/bggdmivvn369gzvt44ta6d2a8ct1l7qn8r.png)
The cube root can be written using a rational exponent.
![\log\left(\sqrt[3]{2-x}}\right) - \log{3x} = \log \left( (2-x)^(1/3) \right) - \log{3x}](https://img.qammunity.org/2022/formulas/mathematics/high-school/m033zoa9dyyo2nwh2gwyc4nqs81zuc2tl0.png)
Use the exponent property to move the exponent 1/3 out.

This now matches answer choice B.