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Question Number 116610 by moh175 last updated on 05/Oct/20

$$$$$$solve:$$$${\left(\frac{1}{\sqrt{2}}\right)}^{2h}+{\left(\frac{\sqrt{3}}{2}\right)}^h=1$$

Commented by Dwaipayan Shikari last updated on 05/Oct/20

$$\mathrm{h}=2\left(\mathrm{By}\mathrm{observation}\right)$$

Answered by 1549442205PVT last updated on 05/Oct/20

$$\mathrm{We}\mathrm{have}$$$${\left(\frac{1}{\sqrt{2}}\right)}^{2h}+{\left(\frac{\sqrt{3}}{2}\right)}^h=1\iff {\left(\frac{1}{2}\right)}^{\mathrm{h}}+{\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{h}}=1\left(1\right)$$$$\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{h}=2\mathrm{satisfy}\mathrm{the}$$$$\mathrm{equation}\mathrm{since}{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1$$$$\mathrm{We}\mathrm{note}\mathrm{that}0<\frac{1}{2}<\frac{\sqrt{3}}{2}<1.\mathrm{Hence}$$$${\left(\frac{1}{2}\right)}^{\mathrm{x}}\mathrm{and}{\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{x}}\mathrm{both}\mathrm{are}\mathrm{decreasing}\mathrm{functions}$$$$\mathrm{on}(0;+\mathrm{\infty }).\mathrm{Therefore}$$$$\mathrm{i})\mathrm{If}\mathrm{h}>2\mathrm{then}{\left(\frac{1}{2}\right)}^{\mathrm{h}}+{\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{h}}<{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=1$$$$\mathrm{ii})\mathrm{If}\mathrm{h}<2\mathrm{then}$$$$\mathrm{then}{\left(\frac{1}{2}\right)}^{\mathrm{h}}+{\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{h}}>{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=1$$$$\mathrm{That}\mathrm{shows}\mathrm{h}=2\mathrm{is}\mathrm{unique}\mathrm{root}\mathrm{of}$$$$\mathrm{given}\mathrm{equation}$$

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