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Question Number 69665 by ozodbek last updated on 26/Sep/19

Commented by ozodbek last updated on 26/Sep/19

$$\mathrm{solve}$$

Answered by mr W last updated on 26/Sep/19

$$x^y\ln x=\ln 9$$$$x^y=\frac{\ln 9}{\ln x}$$$$y\ln x=\ln \left(\frac{\ln 9}{\ln x}\right)=\ln \left(\ln 9\right)-\ln \left(\ln x\right)$$$$\Rightarrow y=\frac{\ln \left(\ln 9\right)-\ln \left(\ln x\right)}{\ln x}$$$$x^y\ln y=\ln 7$$$$\frac{\ln 9}{\ln x}\ln y=\ln 7$$$$\ln y=\frac{\ln 7\ln x}{\ln 9}$$$$\ln \frac{\ln \left(\ln 9\right)-\ln \left(\ln x\right)}{\ln x}=\frac{\ln 7\ln x}{\ln 9}$$$$\ln [\ln \left(\ln 9\right)-\ln \left(\ln x\right)]-\ln \left(\ln x\right)=\left(\frac{\ln 7}{\ln 9}\right)\ln x$$$$witht=\ln x$$$$\ln [\ln \left(\ln 9\right)-\ln t]-\ln t=\left(\frac{\ln 7}{\ln 9}\right)t$$$$\Rightarrow t=\ln x=0.664263$$$$\Rightarrow x=1.943058$$$$\Rightarrow \ln y=0.588285$$$$\Rightarrow y=1.800899$$

Commented by ozodbek last updated on 27/Sep/19

$$\mathrm{thank}\mathrm{you}$$

Commented by otchereabdullai@gmail.com last updated on 29/Sep/19

$$\mathrm{the}\mathrm{great}\mathrm{prof}\mathrm{W}\mathrm{you}\mathrm{are}\mathrm{truely}\mathrm{exceptional}$$

Commented by mr W last updated on 29/Sep/19

$$nicetoseeyoubackinforum,sir!$$

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