3a. Answer:
[tex]\log _{r^2}(x)[/tex]
Applying the properties of logarithms
[tex]\log _{a^b}(x)=\frac{1}{b}\log _a(x)\text{ assume }a\ge0[/tex]
Therefore, in this case:
[tex]=\frac{1}{2}\log _r(x)[/tex]
3b. Given
[tex]\log _2(y)+\log _4(x)+\log _4(2x)=0[/tex]
So, clear for y:
[tex]\begin{gathered} \log _2(y)+\log _4(2x^2)=0 \\ \log _2(y)+\frac{1}{2}\log _2(2x^2)=0 \\ \log _2(y)=-\frac{1}{2}\log _2(2x^2) \\ \log _2(y)=\log _2(\frac{1}{\sqrt[]{2x}}) \\ y=\frac{1}{\sqrt[]{2}}x^{-1} \end{gathered}[/tex]
Answer:
[tex]y=\frac{1}{\sqrt[]{2}}x^{-1}[/tex]
