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Question-69293




Question Number 69293 by A8;15: last updated on 22/Sep/19
Commented by MJS last updated on 22/Sep/19
great, you′ve been faster than me  but also x=0 is a solution
$$\mathrm{great},\:\mathrm{you}'\mathrm{ve}\:\mathrm{been}\:\mathrm{faster}\:\mathrm{than}\:\mathrm{me} \\ $$$$\mathrm{but}\:\mathrm{also}\:{x}=\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Sep/19
Only for this time sir.   You are always greater & faster!
$$\mathcal{O}{nly}\:{for}\:{this}\:{time}\:{sir}.\: \\ $$$${You}\:{are}\:{always}\:{greater}\:\&\:{faster}! \\ $$
Commented by Rasheed.Sindhi last updated on 22/Sep/19
2^x −3^x =(√(6^x −9^x ))  2^x −3^x =(√(2^x .3^x −(3^x )^2 ))  2^x −3^x =(√3^x )(√(2^x −3^x ))  (√(2^x −3^x ))=(√3^x )  2^x =2.3^x   ((2/3))^x =2  xlog((2/3))=log(2)  x=log(2)/log((2/3))
$$\mathrm{2}^{{x}} −\mathrm{3}^{{x}} =\sqrt{\mathrm{6}^{{x}} −\mathrm{9}^{{x}} } \\ $$$$\mathrm{2}^{{x}} −\mathrm{3}^{{x}} =\sqrt{\mathrm{2}^{{x}} .\mathrm{3}^{{x}} −\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{2}^{{x}} −\mathrm{3}^{{x}} =\sqrt{\mathrm{3}^{{x}} }\sqrt{\mathrm{2}^{{x}} −\mathrm{3}^{{x}} } \\ $$$$\sqrt{\mathrm{2}^{{x}} −\mathrm{3}^{{x}} }=\sqrt{\mathrm{3}^{{x}} } \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}.\mathrm{3}^{{x}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{2} \\ $$$${x}\mathrm{log}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{log}\left(\mathrm{2}\right) \\ $$$${x}=\mathrm{log}\left(\mathrm{2}\right)/\mathrm{log}\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$

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