Respuesta :
Given:
[tex]\frac{1}{6}\log _4z+\log _4(z^2-36)-\frac{7}{6}\log _4(z-6)[/tex]To combine these logarithms into a single logarithm, the first thing we need to do is to re-write the expression as a single logarithm with coeffient.
[tex]\frac{1}{6}\log _4z+\log _4(z^2-36)-\frac{7}{6}\log _4(z-6)=\frac{\log _4\sqrt[6]{z}+\log _4(z^2-36)}{\log _4\sqrt[6]{(z-6)^7}}[/tex]Next, we will use the Product Rule to simplify the numerator.
Product Rule states that:
[tex]\log _b(MN)=\log _bM+\log _bN[/tex]Applying it to the numerator:
[tex]\log _4\sqrt[6]{z}+\log _4(z^2-36)=\log _4(\sqrt[6]{z}(z^2-36))[/tex]We will now have a new expression:
[tex]\frac{\log_4(\sqrt[6]{z}(z^2-36))}{\log_4\sqrt[6]{(z-6)^7}}[/tex]Next, we will factor out the logarithm.
[tex]\frac{\log_4(\sqrt[6]{z}(z^2-36))}{\log_4\sqrt[6]{(z-6)^7}}=\log _4(\frac{\sqrt[6]{z}(z^2-36)}{\sqrt[6]{(z-6)^7}})[/tex]Therefore, the final answer would be:
[tex]\log _4(\frac{\sqrt[6]{z}(z^2-36)}{\sqrt[6]{(z-6)^7}})[/tex]