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3-x-4-x-5-x-




Question Number 72190 by oyemi kemewari last updated on 26/Oct/19
3^x +4^x =5^x
$$\mathrm{3}^{\mathrm{x}} +\mathrm{4}^{\mathrm{x}} =\mathrm{5}^{\mathrm{x}} \\ $$
Commented by oyemi kemewari last updated on 26/Oct/19
find x
Commented by Joel578 last updated on 26/Oct/19
x = 2
$${x}\:=\:\mathrm{2} \\ $$
Answered by mind is power last updated on 27/Oct/19
⇔((3/5))^x +((4/5))^x =1  x=2 worcks  f(x)=((3/5))^x +((4/5))^x   f′(x)=ln((3/5)).((3/5))^x +ln((4/5)).((4/5))^x <0  f(2)=1⇒ has only  2 as solution
$$\Leftrightarrow\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{x}} =\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{2}\:\mathrm{worcks} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{x}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{5}}\right).\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{x}} +\mathrm{ln}\left(\frac{\mathrm{4}}{\mathrm{5}}\right).\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{x}} <\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=\mathrm{1}\Rightarrow\:\mathrm{has}\:\mathrm{only}\:\:\mathrm{2}\:\mathrm{as}\:\mathrm{solution} \\ $$
Commented by Rasheed.Sindhi last updated on 27/Oct/19
((3/5))^x +((4/5))^x =1  Assume (3/5)=sin θ⇒(4/5)=cos θ  (sin θ)^x +(cos θ)^x =1  Comparing with the identity  sin^2 θ +cos^2 θ=1,we have   x=2
$$\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{x}} =\mathrm{1} \\ $$$${Assume}\:\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{sin}\:\theta\Rightarrow\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{cos}\:\theta \\ $$$$\left(\mathrm{sin}\:\theta\right)^{{x}} +\left(\mathrm{cos}\:\theta\right)^{{x}} =\mathrm{1} \\ $$$${Comparing}\:{with}\:{the}\:{identity} \\ $$$$\mathrm{sin}^{\mathrm{2}} \theta\:+\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{1},{we}\:{have}\: \\ $$$${x}=\mathrm{2} \\ $$

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