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Question Number 187191 by 073 last updated on 14/Feb/23

Commented by Rasheed.Sindhi last updated on 14/Feb/23

$$You{don}^{\prime }tdeservesolutionbecause$$$$you{don}^{\prime }tgiveanyfeedback.$$

Commented by 073 last updated on 14/Feb/23

$$\mathrm{please}\mathrm{solve}\mathrm{it}$$$$$$

Answered by aba last updated on 14/Feb/23

$$2^{-1}=2^{-1}\ast \mathrm{e}=2^{-1}\ast (2\ast 2^{-1})=2^{-1}\ast \left(\frac{2\times 2^{-1}}{4}\right)=2^{-1}\ast \frac{1}{4}=\frac{2^{-1}\times \frac{1}{4}}{4}=\frac{1}{32}$$

Commented by 073 last updated on 15/Feb/23

Answered by mr W last updated on 15/Feb/23

$$x\ast y=\frac{xy}{4}$$$$x\ast e=\frac{xe}{4}=x\Rightarrow e=4$$$$x\ast x^{-1}=\frac{x\cdot x^{-1}}{4}=e=4$$$$\Rightarrow x\cdot x^{-1}=4\times 4=16\Rightarrow x^{-1}=\frac{16}{x}$$$$\Rightarrow 2^{-1}=\frac{16}{2}=8$$

Commented by 073 last updated on 15/Feb/23

$$\mathrm{thanks}\mathrm{alot}$$

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