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,

\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log 100\] Show Step 3

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

\[{\log _b}\left( {xy} \right) = {\log _b}x + {\log _b}y\hspace{0.25in}\hspace{0.25in}{\log _b}\left( {\frac{x}{y}} \right) = {\log _b}x - {\log _b}y\]

and in fact we’ll need to use both in the end.

The first two logarithms are a difference so let’s use the quotient property to first combine those to get,

\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log {10^2} = \log \left( {\frac{{\sqrt[3]{a}}}{{{b^6}}}} \right) + \log 100\]

We converted the fractional exponent in the first term to a root to make the answer a little nicer but doesn’t really need to be done in general.

Show Step 4

Finally, note that we now have a sum of two logarithms and we can use the product property to combine those to get,

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