Question Number 208463 by hardmath last updated on 16/Jun/24
$$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\mathrm{a}\:=\:\mathrm{log}_{\mathrm{3}} \:\mathrm{4} \\ $$$$\mathrm{b}\:=\:\mathrm{log}_{\mathrm{5}} \:\mathrm{6} \\ $$$$\mathrm{c}\:=\:\mathrm{log}_{\mathrm{6}} \:\mathrm{2} \\ $$
Answered by Frix last updated on 16/Jun/24
$$\mathrm{3}^{{a}} =\mathrm{4}\:\Rightarrow\:{a}>\mathrm{1}\:\:\:\:\:\mathrm{3}^{{a}} =\mathrm{3}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{a}>{b} \\ $$$$\mathrm{5}^{{b}} =\mathrm{6}\:\Rightarrow\:{b}>\mathrm{1}\:\:\:\:\:\:\mathrm{5}^{{b}} =\mathrm{5}+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{6}^{{c}} =\mathrm{2}\:\Rightarrow\:{c}<\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:{a}>{b}>{c} \\ $$
Commented by hardmath last updated on 16/Jun/24
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$