Question Number 145406 by mathdanisur last updated on 04/Jul/21
$${if}\:\:\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}+\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{b}}=\mathrm{2}\:;\:\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}+\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}=\mathrm{0} \\ $$$${find}\:\:\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{b}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{a}}}\:=\:? \\ $$
Answered by som(math1967) last updated on 04/Jul/21
![log_b c+log_a c=0 log_b c×log_c b=−log_a c×log_c b 1=−log_a b ∴log_b a=−1 (1/(log_a b))+(1/(log_b c))+(1/(log_c a)) log_b a+log_c b+log_a c −1+2=1 [∵log_a c+log_c b=2]](Q145413.png)
$$\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}+\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}=\mathrm{0} \\ $$$$\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}×\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{b}}=−\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}×\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{b}} \\ $$$$\mathrm{1}=−\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{b}} \\ $$$$\therefore\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{a}}=−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{b}}}+\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}}+\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{a}}} \\ $$$$\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{a}}+\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{b}}+\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}} \\ $$$$−\mathrm{1}+\mathrm{2}=\mathrm{1}\:\left[\because\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}+\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{b}}=\mathrm{2}\right] \\ $$$$ \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${alot}\:{cool}\:{thankyou}\:{Ser} \\ $$
Answered by liberty last updated on 05/Jul/21
$$\:\begin{cases}{\mathrm{log}\:_{{a}} {c}+\mathrm{log}\:_{{c}} {b}\:=\:\mathrm{2}\:...\left(\mathrm{1}\right)}\\{\mathrm{log}\:_{{b}} {c}\:+\:\mathrm{log}\:_{{a}} {c}\:=\:\mathrm{0}\:...\left(\mathrm{2}\right)}\end{cases} \\ $$$$\:\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{log}\:_{{a}} {c}\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} {a}} \\ $$$${so}\:\mathrm{log}\:_{{b}} {c}\:+\:\mathrm{log}\:_{{a}} {c}\:=\:\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} {b}}\:=\:−\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} {a}}\:\mathrm{or}\:\mathrm{log}\:_{{c}} {a}\:=−\mathrm{log}\:_{{c}} {b}\: \\ $$$$\mathrm{then}\:{a}\:=\:\frac{\mathrm{1}}{{b}}\:...\left(\mathrm{3}\right) \\ $$$$\mathrm{substract}\:\mathrm{eq}\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right)\:\mathrm{give}\: \\ $$$$\Rightarrow\:\mathrm{log}\:_{{c}} {b}−\mathrm{log}\:_{{b}} {c}\:=\:\mathrm{2}\: \\ $$$${now}\:{consider}\:{that}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{log}\:_{{a}} {b}}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{b}} {c}}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} {a}}\:= \\ $$$$\mathrm{log}\:_{{b}} {a}\:+\:\mathrm{log}\:_{{c}} {b}\:+\:\mathrm{log}\:_{{a}} {c}\:= \\ $$$$−\mathrm{1}+\:\mathrm{log}\:_{{c}} {b}−\mathrm{log}\:_{{b}} {c}\:= \\ $$$$−\mathrm{1}+\mathrm{2}\:=\:\mathrm{1}.\: \\ $$
Commented by mathdanisur last updated on 05/Jul/21
$${alot}\:{cool}\:{thanks}\:{Ser} \\ $$