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Question Number 88642 by M±th+et£s last updated on 11/Apr/20

Commented by ajfour last updated on 11/Apr/20

x=(π/4) , of course qualifies. both terms are zero.

$${x}=\frac{\pi}{\mathrm{4}}\:,\:\:{of}\:{course}\:{qualifies}. \\ $$$${both}\:{terms}\:{are}\:{zero}. \\ $$

Commented by M±th+et£s last updated on 11/Apr/20

yes sir but how can we get the value

$${yes}\:{sir}\:{but}\:{how}\:{can}\:{we}\:{get}\:{the}\:{value} \\ $$

Answered by ajfour last updated on 11/Apr/20

((ln tan x)/(ln cos x))+((ln cot x)/(ln sin x))=0 ((ln sin x−ln cos x)/(ln cos x))+((ln cos x−ln sin x)/(ln sin x))=0 ((ln sin x)/(ln cos x))−1+((ln cos x)/(ln sin x))−1=0 let ((ln sin x)/(ln cos x))=t ⇒ t−1+(1/t)−1=0 t+(1/t)=2 ⇒ (t^2 −2t+1)=0 (t−1)^2 =0 ⇒ t=1 ⇒ ((ln sin x)/(ln cos x))=1 ⇒ ln sin x−ln cos x=0 ⇒ ln (((sin x)/(cos x)))=0 tan x=1 ⇒ x=(π/4) . can that do ?

$$\frac{\mathrm{ln}\:\mathrm{tan}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}+\frac{\mathrm{ln}\:\mathrm{cot}\:{x}}{\mathrm{ln}\:\mathrm{sin}\:{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\mathrm{sin}\:{x}−\mathrm{ln}\:\mathrm{cos}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}+\frac{\mathrm{ln}\:\mathrm{cos}\:{x}−\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{sin}\:{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}−\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{cos}\:{x}}{\mathrm{ln}\:\mathrm{sin}\:{x}}−\mathrm{1}=\mathrm{0} \\ $$$${let}\:\:\:\frac{\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}={t} \\ $$$$\Rightarrow\:\:{t}−\mathrm{1}+\frac{\mathrm{1}}{{t}}−\mathrm{1}=\mathrm{0} \\ $$$${t}+\frac{\mathrm{1}}{{t}}=\mathrm{2}\:\:\Rightarrow\:\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\:\:\:\Rightarrow\:\:{t}=\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\mathrm{ln}\:\mathrm{sin}\:{x}}{\mathrm{ln}\:\mathrm{cos}\:{x}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{ln}\:\mathrm{sin}\:{x}−\mathrm{ln}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{ln}\:\left(\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right)=\mathrm{0} \\ $$$$\mathrm{tan}\:{x}=\mathrm{1}\:\:\:\Rightarrow\:\:{x}=\frac{\pi}{\mathrm{4}}\:. \\ $$$${can}\:{that}\:{do}\:? \\ $$

Commented by M±th+et£s last updated on 11/Apr/20

correct but can you explain why log_(cos(x)) tan(x)=((ln(tan(x)))/(ln(cos(x))) and thank you

$${correct} \\ $$$${but}\:{can}\:{you}\:{explain}\:{why}\:{log}_{{cos}\left({x}\right)} {tan}\left({x}\right)=\frac{{ln}\left({tan}\left({x}\right)\right)}{{ln}\left({cos}\left({x}\right)\right.} \\ $$$${and}\:{thank}\:{you}\: \\ $$

Commented by ajfour last updated on 11/Apr/20

basic logarithm property. log_( b) x=t ⇒ x=b^t granted log_( b) x=((log _c x)/(log _c b)) = t ⇒ log _c x=tlog _c b ⇒ log _c x=log _c b^t ⇒ x=b^t . i can just explain this way.

$${basic}\:{logarithm}\:{property}. \\ $$$$\:\:\mathrm{log}_{\:{b}} {x}={t} \\ $$$$\Rightarrow\:\:{x}={b}^{{t}} \: \\ $$$${granted}\:\:\:\mathrm{log}_{\:{b}} {x}=\frac{\mathrm{log}\:_{{c}} {x}}{\mathrm{log}\:_{{c}} {b}}\:=\:{t} \\ $$$$\Rightarrow\:\:\:\mathrm{log}\:_{{c}} {x}={t}\mathrm{log}\:_{{c}} {b} \\ $$$$\Rightarrow\:\:\:\mathrm{log}\:_{{c}} {x}=\mathrm{log}\:_{{c}} {b}^{{t}} \\ $$$$\Rightarrow\:\:\:{x}={b}^{{t}} \:. \\ $$$${i}\:{can}\:{just}\:{explain}\:{this}\:{way}. \\ $$

Commented by M±th+et£s last updated on 12/Apr/20

and i think its (π/4)+2kπ

$${and}\:{i}\:{think}\:{its}\:\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi \\ $$

Commented by M±th+et£s last updated on 12/Apr/20

yes thank you

$${yes}\:{thank}\:{you} \\ $$

Commented by ajfour last updated on 12/Apr/20

yes, i dint care, for i thought, you wd infer that.

$${yes},\:{i}\:{dint}\:{care},\:{for}\:{i}\:{thought}, \\ $$$${you}\:{wd}\:{infer}\:{that}. \\ $$

Commented by M±th+et£s last updated on 12/Apr/20

thank you and god bless you

$${thank}\:{you}\:{and}\:{god}\:{bless}\:{you}\: \\ $$

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