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Question Number 99621 by student work last updated on 22/Jun/20

$$6^{\mathrm{x}}={\mathrm{x}}^5\mathrm{x}=?\mathrm{help}\mathrm{me}$$

Commented by Dwaipayan Shikari last updated on 22/Jun/20

$$\frac{x}{logx}=\frac{5}{log6}$$$$x^{\frac{5}{x}}=6$$$$x=2.199\left(approx\right)$$

Answered by MWSuSon last updated on 22/Jun/20

$${\ln 6}^{\mathrm{x}}={\mathrm{lnx}}^5$$$$\iff \mathrm{xln}6=5\mathrm{lnx}$$$$\iff \frac{1}{5}\ln 6={\mathrm{x}}^{-1}\mathrm{lnx}$$$$\iff \frac{1}{5}\ln 6={\mathrm{e}}^{{\mathrm{lnx}}^{-1}}\mathrm{lnx}$$$$\iff -\frac{1}{5}\ln 6=-{\mathrm{lnxe}}^{-\mathrm{lnx}}$$$${\mathrm{W}}_0(-\frac{1}{5}\ln 6)={\mathrm{W}}_0(-{\mathrm{lnxe}}^{-\mathrm{lnx}})$$$$\mathrm{since}-\frac{1}{5}\ln 6>-\frac{1}{\mathrm{e}}\mathrm{we}\mathrm{use}\mathrm{both}\mathrm{p}-$$$$\mathrm{rincipal}\mathrm{and}-1\mathrm{branch}.$$$$-\mathrm{lnx}={\mathrm{W}}_{\mathrm{p}}(-\frac{1}{5}\ln 6)$$$$\mathrm{lnx}=-{\mathrm{W}}_{\mathrm{p}}(-\frac{1}{5}\ln 6)$$$$\mathrm{x}={\mathrm{e}}^{-{\mathrm{W}}_{\mathrm{p}}(-\frac{1}{5}\ln 6)}$$$$\mathrm{x}=2.1991$$$$\mathrm{Looking}\mathrm{at}\mathrm{the}-1\mathrm{branch}$$$$\mathrm{x}{=}^{-{\mathrm{W}}_{-1}(-\frac{1}{5}\ln 6)}$$$$\mathrm{x}=3.4795$$$$\Rightarrow \mathrm{for}{6}^{\mathrm{x}}={\mathrm{x}}^5$$$$\mathrm{x}=\{\begin{array}{l}2.1991,{\mathrm{W}}_{\mathrm{p}}\\ 3.4795,{\mathrm{W}}_{-1}\end{array}$$$${\mathrm{W}}_0\mathrm{is}\mathrm{the}\mathrm{lambert}\mathrm{W}\mathrm{function}\mathrm{also}$$$$\mathrm{called}\mathrm{product}\log .$$

Answered by mr W last updated on 22/Jun/20

$$x=6^{\frac{x}{5}}$$$$1={xe}^{-\frac{x\ln 6}{5}}$$$$-\frac{\ln 6}{5}=-\frac{x\ln 6}{5}{e}^{-\frac{x\ln 6}{5}}$$$$-\frac{x\ln 6}{5}=W(-\frac{\ln 6}{5})$$$$\Rightarrow x=-\frac{5}{\ln 6}W(-\frac{\ln 6}{5})=\{\begin{array}{l}2.1991\\ 3.4795\end{array}$$

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