Question Number 170491 by jahar last updated on 25/May/22
$$\left.\mathrm{1}\right)\:{Help} \\ $$$$\mathrm{5}^{{x}} −\mathrm{3}^{{x}} =\mathrm{9}\: \\ $$
Answered by MathsFan last updated on 25/May/22
$${let}\:{f}\left({x}\right)=\mathrm{5}^{{x}} −\mathrm{3}^{{x}} −\mathrm{9} \\ $$$$\:{f}\:'\left({x}\right)=\mathrm{5}^{{x}} {ln}\mathrm{5}−\mathrm{3}^{{x}} {ln}\mathrm{3} \\ $$$$\:{let}\:{x}_{\mathrm{0}} =\mathrm{2} \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{0}} −\frac{{f}\left({x}_{\mathrm{0}} \right)}{{f}\:'\left({x}_{\mathrm{0}} \right)}\approx \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} −\frac{{f}\left({x}_{\mathrm{1}} \right)}{{f}\:'\left({x}_{\mathrm{1}} \right)}\approx \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{2}} −\frac{{f}\left({x}_{\mathrm{2}} \right)}{{f}\:'\left({x}_{\mathrm{2}} \right)}\approx \\ $$$${x}_{\mathrm{4}} ={x}_{\mathrm{3}} −\frac{{f}\left({x}_{\mathrm{3}} \right)}{{f}\:'\left({x}_{\mathrm{3}} \right)}\approx \\ $$$$\:\therefore\:{x}\approx{x}_{\mathrm{4}} \\ $$