To convert this into a single logarithm we’ll be using the properties that we used to break up logarithms in reverse. The first step in this process is to use the property,

$${\log _b}\left( {{x^r}} \right) = r{\log _b}x$$

to make sure that all the logarithms have coefficients of one. This needs to be done first because all the properties that allow us to combine sums/differences of logarithms require coefficients of one on individual logarithms. So, using this property gives,

$$\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + 2$$

Now, for the 2 let’s notice that we can write this in terms of a logarithm as,

$$2 = \log {10^2} = \log 100$$

Note that this is really just using the property,

$${\log _b}{b^x} = x$$

So, we now have,

$$\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log 100$$

Now, there are several ways to proceed from this point. We can use either of the two properties.

$${\log _b}\left( {xy} \right) = {\log _b}x + {\log _b}y\hspace{0.75in}{\log _b}\left( {\frac{x}{y}} \right) = {\log _b}x - {\log _b}y$$

and in fact we’ll need to use both in the end. Which we use first does not matter as we’ll end up with the same result in the end. Here is the rest of the work for this problem.

$$\begin{align*}\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log {10^2} & = \log \left( {100\,\sqrt[3]{a}} \right) - \log \left( {{b^6}} \right)\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\log \left( {\frac{{100\,\sqrt[3]{a}}}{{{b^6}}}} \right)}}\end{align*}$$

Note that the only reason we converted the fractional exponent to a root was to make the final answer a little nicer.