log (base 4)28.log (base 3/7)4.log (base 3)3/7 is greater than

a)3
b)4
c)5
​d)6 ​​  

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{L}={\log }_4\left(28\right) {\log }_{\frac{3}{7}}\left(4\right) .{\log }_3\left(\frac{3}{7}\right) \\ \mathrm{Using} \mathrm{base} \mathrm{change} \mathrm{formula} \mathrm{i}.\mathrm{e}. {\log }_{\mathrm{y}}\mathrm{x}=\frac{{\log }_{\mathrm{z}}\mathrm{x}}{{\log }_{\mathrm{z}}\mathrm{y}} \mathrm{we} \mathrm{get} \\ \mathrm{L}={\log }_4\left(28\right) \frac{{\log }_4\left(4\right)}{{\log }_4\left({\displaystyle \frac{3}{7}}\right)}.\frac{{\log }_4\left({\displaystyle \frac{3}{7}}\right)}{{\log }_4\left(3\right)} \\ ={\log }_4\left(28\right) \frac{{\log }_4\left(4\right)}{{\log }_4\left(3\right)} \\ =\frac{{\log }_4\left(28\right)}{{\log }_4\left(3\right)} \left[\mathrm{As} {\log }_4\left(4\right)=1\right] \\ \mathrm{L}={\log }_3\left(28\right) \left[\mathrm{As} {\log }_{\mathrm{y}}\mathrm{x}=\frac{{\log }_{\mathrm{z}}\mathrm{x}}{{\log }_{\mathrm{z}}\mathrm{y}} \right] \\ \mathrm{Now} {\log }_3\left(27\right)={\log }_3\left(3^3\right) \\ =3 {\log }_3\left(3\right) \left[\log \left({\mathrm{m}}^{\mathrm{n}}\right)=\mathrm{n} \log \left(\mathrm{m}\right)\right] \\ \Rightarrow {\log }_3\left(27\right)=3 \\ \Rightarrow {\log }_3\left(28\right)>3...\mathrm{Option}\left(\mathrm{a}\right) \end{gathered}$$

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