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Question Number 145406 by mathdanisur last updated on 04/Jul/21

$$if{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{c}+{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{b}=2;{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{c}+{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{c}=0$$$$find\frac{1}{{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{b}}+\frac{1}{{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{c}}+\frac{1}{{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{a}}=?$$

Answered by som(math1967) last updated on 04/Jul/21

$${\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{c}+{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{c}=0$$$${\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{c}\times {\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{b}=-{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{c}\times {\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{b}$$$$1=-{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{b}$$$$\therefore {\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{a}=-1$$$$\frac{1}{{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{b}}+\frac{1}{{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{c}}+\frac{1}{{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{a}}$$$${\mathit{l}\mathit{o}\mathit{g}}_{\mathit{b}}\mathit{a}+{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{b}+{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{c}$$$$-1+2=1[\because {\mathit{l}\mathit{o}\mathit{g}}_{\mathit{a}}\mathit{c}+{\mathit{l}\mathit{o}\mathit{g}}_{\mathit{c}}\mathit{b}=2]$$$$$$

Commented by mathdanisur last updated on 04/Jul/21

$$alotcoolthankyouSer$$

Answered by liberty last updated on 05/Jul/21

$$\{\begin{array}{l}\log {}_{a}c+\log {}_{c}b=2...\left(1\right)\\ \log {}_{b}c+\log {}_{a}c=0...\left(2\right)\end{array}$$$$\mathrm{we}\mathrm{know}\mathrm{that}\log {}_{a}c=\frac{1}{\log {}_{c}a}$$$$so\log {}_{b}c+\log {}_{a}c=0$$$$\Rightarrow \frac{1}{\log {}_{c}b}=-\frac{1}{\log {}_{c}a}\mathrm{or}\log {}_{c}a=-\log {}_{c}b$$$$\mathrm{then}a=\frac{1}{b}...\left(3\right)$$$$\mathrm{substract}\mathrm{eq}\left(1\right)\mathrm{\&}\left(2\right)\mathrm{give}$$$$\Rightarrow \log {}_{c}b-\log {}_{b}c=2$$$$nowconsiderthat$$$$\frac{1}{\log {}_{a}b}+\frac{1}{\log {}_{b}c}+\frac{1}{\log {}_{c}a}=$$$$\log {}_{b}a+\log {}_{c}b+\log {}_{a}c=$$$$-1+\log {}_{c}b-\log {}_{b}c=$$$$-1+2=1.$$

Commented by mathdanisur last updated on 05/Jul/21

$$alotcoolthanksSer$$

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