Question Number 88642 by M±th+et£s last updated on 11/Apr/20
Commented by ajfour last updated on 11/Apr/20
$${x}=\frac{\pi}{\mathrm{4}}\:,\:\:{of}\:{course}\:{qualifies}. \\ $$$${both}\:{terms}\:{are}\:{zero}. \\ $$
Commented by M±th+et£s last updated on 11/Apr/20
$${yes}\:{sir}\:{but}\:{how}\:{can}\:{we}\:{get}\:{the}\:{value} \\ $$
Answered by ajfour last updated on 11/Apr/20
$$\frac{\mathrm{ln}\:\mathrm{tan}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}+\frac{\mathrm{ln}\:\mathrm{cot}\:{x}}{\mathrm{ln}\:\mathrm{sin}\:{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\mathrm{sin}\:{x}−\mathrm{ln}\:\mathrm{cos}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}+\frac{\mathrm{ln}\:\mathrm{cos}\:{x}−\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{sin}\:{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}−\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{cos}\:{x}}{\mathrm{ln}\:\mathrm{sin}\:{x}}−\mathrm{1}=\mathrm{0} \\ $$$${let}\:\:\:\frac{\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}={t} \\ $$$$\Rightarrow\:\:{t}−\mathrm{1}+\frac{\mathrm{1}}{{t}}−\mathrm{1}=\mathrm{0} \\ $$$${t}+\frac{\mathrm{1}}{{t}}=\mathrm{2}\:\:\Rightarrow\:\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\:\:\:\Rightarrow\:\:{t}=\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{ln}\:\mathrm{sin}\:{x}−\mathrm{ln}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{ln}\:\left(\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right)=\mathrm{0} \\ $$$$\mathrm{tan}\:{x}=\mathrm{1}\:\:\:\Rightarrow\:\:{x}=\frac{\pi}{\mathrm{4}}\:. \\ $$$${can}\:{that}\:{do}\:? \\ $$
Commented by M±th+et£s last updated on 11/Apr/20
$${correct} \\ $$$${but}\:{can}\:{you}\:{explain}\:{why}\:{log}_{{cos}\left({x}\right)} {tan}\left({x}\right)=\frac{{ln}\left({tan}\left({x}\right)\right)}{{ln}\left({cos}\left({x}\right)\right.} \\ $$$${and}\:{thank}\:{you}\: \\ $$
Commented by ajfour last updated on 11/Apr/20
$${basic}\:{logarithm}\:{property}. \\ $$$$\:\:\mathrm{log}_{\:{b}} {x}={t} \\ $$$$\Rightarrow\:\:{x}={b}^{{t}} \: \\ $$$${granted}\:\:\:\mathrm{log}_{\:{b}} {x}=\frac{\mathrm{log}\:_{{c}} {x}}{\mathrm{log}\:_{{c}} {b}}\:=\:{t} \\ $$$$\Rightarrow\:\:\:\mathrm{log}\:_{{c}} {x}={t}\mathrm{log}\:_{{c}} {b} \\ $$$$\Rightarrow\:\:\:\mathrm{log}\:_{{c}} {x}=\mathrm{log}\:_{{c}} {b}^{{t}} \\ $$$$\Rightarrow\:\:\:{x}={b}^{{t}} \:. \\ $$$${i}\:{can}\:{just}\:{explain}\:{this}\:{way}. \\ $$
Commented by M±th+et£s last updated on 12/Apr/20
$${and}\:{i}\:{think}\:{its}\:\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi \\ $$
Commented by M±th+et£s last updated on 12/Apr/20
$${yes}\:{thank}\:{you} \\ $$
Commented by ajfour last updated on 12/Apr/20
$${yes},\:{i}\:{dint}\:{care},\:{for}\:{i}\:{thought}, \\ $$$${you}\:{wd}\:{infer}\:{that}. \\ $$
Commented by M±th+et£s last updated on 12/Apr/20
$${thank}\:{you}\:{and}\:{god}\:{bless}\:{you}\: \\ $$