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Question Number 187336 by Humble last updated on 16/Feb/23

$$solveforxif$$$$X^x\bullet 5^x-5^{2+x}=0$$$$$$

Answered by mr W last updated on 16/Feb/23

$$x^x\times 5^x-25\times 5^x=0$$$$(x^x-25)5^x=0$$$$5^x\ne 0$$$$\Rightarrow x^x-25=0$$$$\Rightarrow x^x=25$$$$\Rightarrow x\ln x=\ln 25$$$$\Rightarrow \left(\ln x\right)e^{\ln x}=\ln 25$$$$\Rightarrow \ln x=W\left(\ln 25\right)$$$$\Rightarrow x=e^{W\left(\ln 25\right)}=\frac{\ln 25}{W\left(\ln 25\right)}$$$$\approx \frac{\ln 25}{1.086276}\approx 2.96322✓$$

Commented by Humble last updated on 16/Feb/23

$$Excellentsolution!thankyou,sir$$

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