Secondaire 5 • 2a
hello
can someone please help me with this question. I can’t find anything on alloprof about how to do it
thanks
hello
can someone please help me with this question. I can’t find anything on alloprof about how to do it
thanks
Explication d'Alloprof
Cette explication a été donnée par un membre de l'équipe d'Alloprof.
Hello PhilosophicalGalaxy9491,
In order to write the expressions as a single logarithm, you can use the logarithm laws.
$$ \begin{align} 4 \ln 2x+ \ln \left( \frac{6}{x} \right)-2\ln 2x &= 4 \ln 2x - 2\ln2x + \ln \left( \frac{6}{x} \right) \, \color{blue}{\text{Regroup similar terms} }\\ &= 2\ln2x + \ln \left( \frac{6}{x} \right)\, \color{blue}{\text{Simplify} } \\ &=\ln(2x)^2 + \ln \left( \frac{6}{x} \right)\, \color{blue}{\text{Power Law for Logarithms }} \\ &=\ln\left((2x)^2 \cdot \frac{6}{x}\right) \, \color{blue}{\text{Product Law for Logarithms}} \\ &= \ln 24x \, \color{blue}{\text{Simplify}} \\ \end{align} $$
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