Section 6.2 : Logarithm Functions
18. Combine \(\displaystyle \frac{1}{3}\log a - 6\log b + 2\) into a single logarithm with a coefficient of one.
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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,
\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + 2\] Show Step 2Now, 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 3Now, 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 4Finally, note that we now have a sum of two logarithms and we can use the product property to combine those to get,
\[\log \left( {{a^{\frac{1}{3}}}} \right) - \log \left( {{b^6}} \right) + \log 100 = \require{bbox} \bbox[2pt,border:1px solid black]{{\log \left( {\frac{{100\,\,\sqrt[3]{a}}}{{{b^6}}}} \right)}}\]