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Question-88642




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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