Since the argument of our initial logarithm is a fraction/division we will use the second property to split the numerator and denominator into separate logarithms:
This argument of the first logarithm is a product so we can use the first property to split the factors into separate logarithms:
Since 1/3 as an exponent means the same thing as a cube root, I am going to replace the cube root with an exponent of 1/3:
Now we can use the third property on the two logarithms with exponents on their arguments. (This is why I changed the cube root into a power of 1/3.)
Last of all, since the first logarithm simplifies to 5:
This, I believe, is the desired answer.