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Question Number 119679 by Khalmohmmad last updated on 26/Oct/20

$$2^1\times 2^2\times 2^3\times ...\times 2^n={(0.5)}^{-45}$$$$n=?$$

Answered by Ar Brandon last updated on 26/Oct/20

$$2^1\times 2^2\times 2^3\times ...\times 2^{\mathrm{n}}={(0.5)}^{-45}$$$$2^{1+2+3+\cdot \cdot \cdot +\mathrm{n}}={\left(2^{-1}\right)}^{-45}=2^{45}$$$$\frac{\mathrm{n}(\mathrm{n}+1)}{2}=45\Rightarrow {\mathrm{n}}^2+\mathrm{n}-90=0$$$$\Rightarrow (\mathrm{n}+10)(\mathrm{n}-9)=0,$$$$\mathrm{n}=-10\vee \mathrm{n}=9$$$$\mathrm{n}\in {\mathbb{Z}}^{+}\Rightarrow \mathrm{n}=9$$

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