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Question Number 170491 by jahar last updated on 25/May/22

1) Help  5^x −3^x =9

$$\left.\mathrm{1}\right)\:{Help} \\ $$$$\mathrm{5}^{{x}} −\mathrm{3}^{{x}} =\mathrm{9}\: \\ $$

Answered by MathsFan last updated on 25/May/22

let f(x)=5^x −3^x −9   f ′(x)=5^x ln5−3^x ln3   let x_0 =2  x_1 =x_0 −((f(x_0 ))/(f ′(x_0 )))≈  x_2 =x_1 −((f(x_1 ))/(f ′(x_1 )))≈  x_3 =x_2 −((f(x_2 ))/(f ′(x_2 )))≈  x_4 =x_3 −((f(x_3 ))/(f ′(x_3 )))≈   ∴ x≈x_4

$${let}\:{f}\left({x}\right)=\mathrm{5}^{{x}} −\mathrm{3}^{{x}} −\mathrm{9} \\ $$$$\:{f}\:'\left({x}\right)=\mathrm{5}^{{x}} {ln}\mathrm{5}−\mathrm{3}^{{x}} {ln}\mathrm{3} \\ $$$$\:{let}\:{x}_{\mathrm{0}} =\mathrm{2} \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{0}} −\frac{{f}\left({x}_{\mathrm{0}} \right)}{{f}\:'\left({x}_{\mathrm{0}} \right)}\approx \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} −\frac{{f}\left({x}_{\mathrm{1}} \right)}{{f}\:'\left({x}_{\mathrm{1}} \right)}\approx \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{2}} −\frac{{f}\left({x}_{\mathrm{2}} \right)}{{f}\:'\left({x}_{\mathrm{2}} \right)}\approx \\ $$$${x}_{\mathrm{4}} ={x}_{\mathrm{3}} −\frac{{f}\left({x}_{\mathrm{3}} \right)}{{f}\:'\left({x}_{\mathrm{3}} \right)}\approx \\ $$$$\:\therefore\:{x}\approx{x}_{\mathrm{4}} \\ $$

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