I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.4 : Simplifying Logarithms
Simplify each of the following logarithms. Show All Solutions Hide All Solutions
- \(\ln {x^3}{y^4}{z^5}\)
Show SolutionHere simplify means use Property 1 - 7 from the Logarithm Properties section as often as you can. You will be done when you can’t use any more of these properties.
Property 5 can be extended to products of more than two functions so,
\[\begin{align*}\ln {x^3}{y^4}{z^5} & = \ln {x^3} + \ln {y^4} + \ln {z^5}\\ & = 3\ln x + 4\ln y + 5\ln z\end{align*}\] - \({\log _3}\left( {\frac{{9{x^4}}}{{\sqrt y }}} \right)\)
Show SolutionIn using property 6 make sure that the logarithm that you subtract is the one that contains the denominator as its argument. Also, note that I’ll be converting the root to exponents in the first step since we’ll need that done for a later step.
\[\begin{align*}{\log _3}\left( {\frac{{9{x^4}}}{{\sqrt y }}} \right) & = {\log _3}9{x^4} - {\log _3}{y^{\frac{1}{2}}}\\ & = {\log _3}9 + {\log _3}{x^4} - {\log _3}{y^{\frac{1}{2}}}\\ & = 2 + 4{\log _3}x - \frac{1}{2}{\log _3}y\end{align*}\]Evaluate logs where possible as I did in the first term.
- \(\log \left( {\frac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^3}}}} \right)\)
Show SolutionThe point to this problem is mostly the correct use of property 7.
\[\begin{align*}\log \left( {\frac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^3}}}} \right) & = \log \left( {{x^2} + {y^2}} \right) - \log {\left( {x - y} \right)^3}\\ & = \log \left( {{x^2} + {y^2}} \right) - 3\log \left( {x - y} \right)\end{align*}\]You can use Property 7 on the second term because the WHOLE term was raised to the 3, but in the first logarithm, only the individual terms were squared and not the term as a whole so the 2’s must stay where they are!