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,

\[\ln {\left( {t + 5} \right)^3} - \ln \left( {{t^4}} \right) - \ln {\left( {s - 1} \right)^2}\]

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.

We should also be careful with the fact that there are two minus signs in here as that sometimes adds confusion to the problem. They are easy to deal with however if we just factor a minus sign out of the last two terms and then proceed from there as follows.

\[\begin{align*}3\ln \left( {t + 5} \right) - 4\ln t - 2\ln \left( {s - 1} \right) &= \ln {\left( {t + 5} \right)^3} - \left( {\ln \left( {{t^4}} \right) + \ln {{\left( {s - 1} \right)}^2}} \right)\\ & = \ln {\left( {t + 5} \right)^3} - \ln \left( {{t^4}{{\left( {s - 1} \right)}^2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\ln \frac{{{{\left( {t + 5} \right)}^3}}}{{{t^4}{{\left( {s - 1} \right)}^2}}}}}\end{align*}\]