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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

$$x=\frac{\pi }{4},ofcoursequalifies.$$$$bothtermsarezero.$$

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

$$yessirbuthowcanwegetthevalue$$

Answered by ajfour last updated on 11/Apr/20

$$\frac{\ln \tan x}{\ln \cos x}+\frac{\ln \cot x}{\ln \sin x}=0$$$$\frac{\ln \sin x-\ln \cos x}{\ln \cos x}+\frac{\ln \cos x-\ln \sin x}{\ln \sin x}=0$$$$\frac{\ln \sin x}{\ln \cos x}-1+\frac{\ln \cos x}{\ln \sin x}-1=0$$$$let\frac{\ln \sin x}{\ln \cos x}=t$$$$\Rightarrow t-1+\frac{1}{t}-1=0$$$$t+\frac{1}{t}=2\Rightarrow (t^2-2t+1)=0$$$${(t-1)}^2=0\Rightarrow t=1$$$$\Rightarrow \frac{\ln \sin x}{\ln \cos x}=1$$$$\Rightarrow \ln \sin x-\ln \cos x=0$$$$\Rightarrow \ln \left(\frac{\sin x}{\cos x}\right)=0$$$$\tan x=1\Rightarrow x=\frac{\pi }{4}.$$$$canthatdo?$$

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

$$correct$$$$butcanyouexplainwhy{log}_{cos\left(x\right)}tan\left(x\right)=\frac{ln\left(tan\left(x\right)\right)}{ln(cos\left(x\right)}$$$$andthankyou$$

Commented by ajfour last updated on 11/Apr/20

$$basiclogarithmproperty.$$$${\log }_{b}x=t$$$$\Rightarrow x=b^t$$$$granted{\log }_{b}x=\frac{\log {}_{c}x}{\log {}_{c}b}=t$$$$\Rightarrow \log {}_{c}x=t\log {}_{c}b$$$$\Rightarrow \log {}_{c}x=\log {}_c{b}^t$$$$\Rightarrow x=b^t.$$$$icanjustexplainthisway.$$

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

$$andithinkits\frac{\pi }{4}+2k\pi$$

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

$$yesthankyou$$

Commented by ajfour last updated on 12/Apr/20

$$yes,idintcare,forithought,$$$$youwdinferthat.$$

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

$$thankyouandgodblessyou$$

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