Question Number 25315 by ibraheem160 last updated on 08/Dec/17
$${if}\:{log}_{{a}\:} ^{{b}} ={log}_{{b}} ^{{c}} ={log}_{{c}} ^{{a}} .\:{show}\:{that}\:{a}={b}={c} \\ $$
Answered by prakash jain last updated on 08/Dec/17
$$\mathrm{log}_{{a}} {b}\mathrm{log}_{{b}} {c}=\mathrm{log}_{{a}} {c} \\ $$$$\Rightarrow\left(\mathrm{log}_{{c}} {a}\right)^{\mathrm{2}} =\mathrm{log}_{{a}} {c} \\ $$$$\Rightarrow\mathrm{1}=\left(\mathrm{log}_{{a}} {c}\right)^{\mathrm{3}} \Rightarrow\mathrm{log}_{{a}} {c}=\mathrm{1} \\ $$$$\Rightarrow{a}={c} \\ $$$$\mathrm{log}_{{c}} {a}=\mathrm{log}_{{b}} {c}=\mathrm{log}_{{a}} {b}=\mathrm{1} \\ $$$$\Rightarrow{a}={b}={c} \\ $$
Answered by Rasheed.Sindhi last updated on 08/Dec/17
$$\:\:\mathrm{Let}\:\mathrm{log}_{\mathrm{a}} \mathrm{b}=\mathrm{log}_{\mathrm{b}} \mathrm{c}=\mathrm{log}_{\mathrm{c}} \mathrm{a}=\mathrm{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{y}} =\mathrm{b}\:,\:\mathrm{b}^{\mathrm{y}} =\mathrm{c}\:,\:\mathrm{c}^{\mathrm{y}} =\mathrm{a} \\ $$$$\:\mathrm{a}^{\mathrm{y}} =\mathrm{b}\:,\:\mathrm{b}^{\mathrm{y}} =\mathrm{c}\Rightarrow\left(\mathrm{a}^{\mathrm{y}} \right)^{\mathrm{y}} =\mathrm{c} \\ $$$$\:\mathrm{c}^{\mathrm{y}} =\mathrm{a}\:,\:\left(\mathrm{a}^{\mathrm{y}} \right)^{\mathrm{y}} =\mathrm{c}\Rightarrow\left\{\left(\mathrm{c}^{\mathrm{y}} \right)^{\mathrm{y}} \right\}^{\mathrm{y}} =\mathrm{c} \\ $$$$\:\mathrm{c}^{\mathrm{y}^{\mathrm{3}} } =\mathrm{c}\Rightarrow\mathrm{y}^{\mathrm{3}} =\mathrm{1}\Rightarrow\mathrm{y}=\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{y}} =\mathrm{b}\Rightarrow\mathrm{a}^{\mathrm{1}} =\mathrm{b}\Rightarrow\mathrm{a}=\mathrm{b} \\ $$$$\mathrm{b}^{\mathrm{y}} =\mathrm{c}\Rightarrow\mathrm{b}^{\mathrm{1}} =\mathrm{c}\Rightarrow\mathrm{b}=\mathrm{c} \\ $$$$\mathrm{a}=\mathrm{b},\mathrm{b}=\mathrm{c}\Rightarrow\mathrm{a}=\mathrm{b}=\mathrm{c} \\ $$