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

The number of real solutions of 3^x +4^x =5^x is ____.

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

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

let f(x) = 3^x +4^x −5^x = 0 for x ≤2 , f(x) is positive. for x x ≥ 3, f(x) is negative. so f(x) is a decreasing function thus it must have only one root between 2 and 3. x ≈ 2.73...

$$\mathrm{let}\:{f}\left({x}\right)\:=\:\mathrm{3}^{{x}} +\mathrm{4}^{{x}} \:−\mathrm{5}^{{x}} \:=\:\mathrm{0} \\ $$$$\mathrm{for}\:{x}\:\leqslant\mathrm{2}\:,\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{positive}. \\ $$$$\mathrm{for}\:{x}\:{x}\:\geqslant\:\mathrm{3},\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{negative}. \\ $$$$\mathrm{so}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{decreasing}\:\mathrm{function}\:\mathrm{thus} \\ $$$$\mathrm{it}\:\mathrm{must}\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{root}\:\mathrm{between}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}. \\ $$$${x}\:\approx\:\mathrm{2}.\mathrm{73}... \\ $$

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

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

$${x}=\mathrm{2}: \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$

Answered by 1549442205 last updated on 12/Jun/20

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

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

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