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Question Number 98057 by PengagumRahasiamu last updated on 11/Jun/20

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{real}\mathrm{solutions}\mathrm{of}$$$$3^x+4^x=5^x\mathrm{is}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Answered by Rio Michael last updated on 11/Jun/20

$$\mathrm{let}f\left(x\right)=3^x+4^x-5^x=0$$$$\mathrm{for}x⩽2,f\left(x\right)\mathrm{is}\mathrm{positive}.$$$$\mathrm{for}xx⩾3,f\left(x\right)\mathrm{is}\mathrm{negative}.$$$$\mathrm{so}f\left(x\right)\mathrm{is}\mathrm{a}\mathrm{decreasing}\mathrm{function}\mathrm{thus}$$$$\mathrm{it}\mathrm{must}\mathrm{have}\mathrm{only}\mathrm{one}\mathrm{root}\mathrm{between}2\mathrm{and}3.$$$$x\approx 2.73...$$

Commented by mr W last updated on 11/Jun/20

$$x=2:$$$$3^2+4^2=5^2$$

Answered by 1549442205 last updated on 12/Jun/20

$$\mathrm{We}\mathrm{prove}\mathrm{that}\mathrm{this}\mathrm{equation}\mathrm{has}\mathrm{unique}$$$$\mathrm{real}\mathrm{root}\mathrm{x}=2.\mathrm{Indeed},\mathrm{dividing}\mathrm{both}\mathrm{two}\mathrm{sides}\mathrm{of}$$$$\mathrm{given}\mathrm{equation}\mathrm{by}{5}^{\mathrm{x}}\mathrm{we}\mathrm{get}$$$${\left(\frac{3}{5}\right)}^{\mathrm{x}}+{\left(\frac{5}{5}\right)}^{\mathrm{x}}=1.\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{x}=2\mathrm{satisfy}\mathrm{since}{\left(\frac{3}{5}\right)}^2+{\left(\frac{5}{5}\right)}^2=\frac{9}{25}+\frac{16}{25}=1$$$$\mathrm{which}\mathrm{means}\mathrm{that}\mathrm{x}=2\mathrm{be}\mathrm{a}\mathrm{root}\mathrm{of}\mathrm{given}\mathrm{eq}.$$$$\mathrm{Also},\mathrm{it}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{functions}$$$$\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{3}{5}\right)}^{\mathrm{x}}\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)={\left(\frac{4}{5}\right)}^{\mathrm{x}}\mathrm{are}\mathrm{decreasing}\mathrm{on}(-\mathrm{\infty };+\mathrm{\infty })$$$$\mathrm{because}\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)={\left(\frac{3}{5}\right)}^{\mathrm{x}}\ln \frac{3}{5}<0,{\mathrm{g}}^{\prime }\left(\mathrm{x}\right)={\left(\frac{4}{5}\right)}^{\mathrm{x}}\ln \frac{4}{5}<0$$$$\mathrm{on}(-\mathrm{\infty };+\mathrm{\infty }).\mathrm{Hence},$$$$\mathrm{a}/\mathrm{For}\mathrm{x}<2\mathrm{we}\mathrm{have}{\left(\frac{3}{5}\right)}^{\mathrm{x}}+{\left(\frac{5}{5}\right)}^{\mathrm{x}}>{\left(\frac{3}{5}\right)}^2+{\left(\frac{5}{5}\right)}^2=1$$$$\mathrm{b}/\mathrm{For}\mathrm{x}>2\mathrm{we}\mathrm{have}{\left(\frac{3}{5}\right)}^{\mathrm{x}}+{\left(\frac{5}{5}\right)}^{\mathrm{x}}<{\left(\frac{3}{5}\right)}^2+{\left(\frac{5}{5}\right)}^2=1$$$$\mathrm{That}\mathrm{show}\mathrm{that}\mathrm{x}=2\mathrm{is}\mathrm{unique}\mathrm{real}\mathrm{root}\mathrm{of}$$$$\mathrm{the}\mathrm{given}\mathrm{equation}$$

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