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.
