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Question Number 69293 by A8;15: last updated on 22/Sep/19

Commented by MJS last updated on 22/Sep/19

$$\mathrm{great},{\mathrm{you}}^{\prime }\mathrm{ve}\mathrm{been}\mathrm{faster}\mathrm{than}\mathrm{me}$$$$\mathrm{but}\mathrm{also}x=0\mathrm{is}\mathrm{a}\mathrm{solution}$$

Commented by Rasheed.Sindhi last updated on 22/Sep/19

$$\mathcal{O}nlyforthistimesir.$$$$Youarealwaysgreater\mathrm{\&}faster!$$

Commented by Rasheed.Sindhi last updated on 22/Sep/19

$$2^x-3^x=\sqrt{6^x-9^x}$$$$2^x-3^x=\sqrt{2^x.3^x-{\left(3^x\right)}^2}$$$$2^x-3^x=\sqrt{3^x}\sqrt{2^x-3^x}$$$$\sqrt{2^x-3^x}=\sqrt{3^x}$$$$2^x=2.3^x$$$${\left(\frac{2}{3}\right)}^x=2$$$$x\log \left(\frac{2}{3}\right)=\log \left(2\right)$$$$x=\log \left(2\right)/\log \left(\frac{2}{3}\right)$$

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