Question
Show that :$\log_a m \div \log_{ab} m + 1 + \log _ab$
✓
Answer
$ \log _{\mathrm{a}} \mathrm{m} \div \log _{\mathrm{ab}} \mathrm{m}==\frac{\log _a m}{\log _{a b} m}$
$ =\frac{\log _m a b}{\log _m a}\left[Q \log _b a=\frac{1}{\log _a b}\right]$
$ =\log _{\mathrm{a}} \mathrm{ab}\left[Q \frac{\log _x a}{\log _{\mathrm{x}} b}=\log _b a\right]$
$ =\log _{\mathrm{a}} \mathrm{a}+\log _{\mathrm{a}} \mathrm{b}$
$ =1+\log _{\mathrm{a}} \mathrm{b}$
$ =\frac{\log _m a b}{\log _m a}\left[Q \log _b a=\frac{1}{\log _a b}\right]$
$ =\log _{\mathrm{a}} \mathrm{ab}\left[Q \frac{\log _x a}{\log _{\mathrm{x}} b}=\log _b a\right]$
$ =\log _{\mathrm{a}} \mathrm{a}+\log _{\mathrm{a}} \mathrm{b}$
$ =1+\log _{\mathrm{a}} \mathrm{b}$
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