Question Number 119802 by bemath last updated on 27/Oct/20
$${Given}\:{a},{b},{c}\:{real}\:{number}\:{and}\:{not}\:{equal}\:{to}\:\mathrm{1}. \\ $$$${If}\:\mathrm{log}\:_{{a}} \left({b}\right)+\mathrm{log}\:_{{b}} \left({c}\right)+\mathrm{log}\:_{{c}} \left({a}\right)=\mathrm{0}\:{then}\: \\ $$$$\left(\mathrm{log}\:_{{a}} \left({b}\right)\right)^{\mathrm{3}} +\left(\mathrm{log}\:_{{b}} \left({c}\right)\right)^{\mathrm{3}} +\left(\mathrm{log}\:_{{c}} \left({a}\right)\right)^{\mathrm{3}} =? \\ $$
Answered by $@y@m last updated on 27/Oct/20
$${If}\:{l}+{m}+{n}=\mathrm{0} \\ $$$${then}\:{l}^{\mathrm{3}} +{m}^{\mathrm{3}} +{n}^{\mathrm{3}} =\mathrm{3}{lmn} \\ $$$${Here}\:{l}=\mathrm{log}\:_{{a}} \left({b}\right) \\ $$$${m}=\mathrm{log}\:_{{b}} \left({c}\right) \\ $$$${n}=\mathrm{log}\:_{{c}} \left({a}\right) \\ $$$$\therefore\:\mathrm{3}{lmn}=\mathrm{3}×\mathrm{log}\:_{{a}} \left({b}\right)×\mathrm{log}\:_{{b}} \left({c}\right)×\mathrm{log}\:_{{c}} \left({a}\right)=\mathrm{3} \\ $$
Answered by Ar Brandon last updated on 27/Oct/20
![log_a b+log_b c+log_c a=0 ⇒(log_a b+log_b c)^3 =−(log_c a)^3 ⇒(log_a b)^3 +3(log_a b)^2 (log_b c) +3(log_a b)(log_b c)^2 +(log_b c)^3 =−(log_c a)^3 ⇒(log_c a)^3 +(log_a b)^3 +(log_b c)^3 =−3(log_a b)(log_b c)[log_a b+log_b c] =3(log_c a)(log_a b)(log_b c) =3((1/(log_a c)))(((log_a b)/1))(((log_a c)/(log_a b)))=3](https://www.tinkutara.com/question/Q119804.png)
$$\mathrm{log}_{\mathrm{a}} \mathrm{b}+\mathrm{log}_{\mathrm{b}} \mathrm{c}+\mathrm{log}_{\mathrm{c}} \mathrm{a}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}+\mathrm{log}_{\mathrm{b}} \mathrm{c}\right)^{\mathrm{3}} =−\left(\mathrm{log}_{\mathrm{c}} \mathrm{a}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}\right)^{\mathrm{2}} \left(\mathrm{log}_{\mathrm{b}} \mathrm{c}\right) \\ $$$$+\mathrm{3}\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}\right)\left(\mathrm{log}_{\mathrm{b}} \mathrm{c}\right)^{\mathrm{2}} +\left(\mathrm{log}_{\mathrm{b}} \mathrm{c}\right)^{\mathrm{3}} =−\left(\mathrm{log}_{\mathrm{c}} \mathrm{a}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\left(\mathrm{log}_{\mathrm{c}} \mathrm{a}\right)^{\mathrm{3}} +\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}\right)^{\mathrm{3}} +\left(\mathrm{log}_{\mathrm{b}} \mathrm{c}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:=−\mathrm{3}\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}\right)\left(\mathrm{log}_{\mathrm{b}} \mathrm{c}\right)\left[\mathrm{log}_{\mathrm{a}} \mathrm{b}+\mathrm{log}_{\mathrm{b}} \mathrm{c}\right] \\ $$$$\:\:\:\:\:=\mathrm{3}\left(\mathrm{log}_{\mathrm{c}} \mathrm{a}\right)\left(\mathrm{log}_{\mathrm{a}} \mathrm{b}\right)\left(\mathrm{log}_{\mathrm{b}} \mathrm{c}\right) \\ $$$$\:\:\:\:\:=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{a}} \mathrm{c}}\right)\left(\frac{\mathrm{log}_{\mathrm{a}} \mathrm{b}}{\mathrm{1}}\right)\left(\frac{\mathrm{log}_{\mathrm{a}} \mathrm{c}}{\mathrm{log}_{\mathrm{a}} \mathrm{b}}\right)=\mathrm{3} \\ $$