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solve for x using log or any other method you wish.

2006-12-17 17:48:44 · 4 answers · asked by starstrucktv 2 in Science & Mathematics Mathematics

to syikin_310: nope i do not know the answer, i've tried to solve it but still can't get it that's why i posted this

2006-12-17 18:12:47 · update #1

4 answers

let y=x^6-2*x^5+0.98; y’=6x^5-10x^4=0, x1=5/3 & y(5/3)=-3.31, x2=0 & y(0)=0.98; thus local and abs min is point [5/3,-3.31], which means that y(x) has 2 roots only; one of them is close to 1, see! y(1)=-0.02, while and y’(1)=-4;
now linear approximation dy=y’(1)*dx or 0.02=-4*dx and dx=-0.005 so x1=1+dx=0.995 and y(x1)=-0.00012;
to find the second root x2 I’d prefer Excel and iterations: next x=2-0.98/x^5 starting with x=2; thus
2 ~1.969375 ~1.966918593 ~1.966711506 ~1.966693977 ~1.966692492 =1.966694

2006-12-18 08:01:04 · answer #1 · answered by Anonymous · 0 0

Transpose everything to the left
x^6 - 2x^5 + 0.98 = 0

Multiply 50 to both sides
50x^6 - 100x^5 + 49 = 0

This is a "6th degree polynomial" equation, and according to Abel-Ruffini Theorem there is no general solutions to such equations. If you want then use very complicated tools in answering such equations, but it would not be very practical, unless you are setting up the equation.

^_^

2006-12-17 18:00:25 · answer #2 · answered by kevin! 5 · 0 1

Newton's iterative method seems appropriate here.
The two real roots are approximately :
0.9949683588 and 1.9666923542.

2006-12-17 20:53:53 · answer #3 · answered by falzoon 7 · 1 0

I think u know it already rite?why did u ask huh????

2006-12-17 18:03:38 · answer #4 · answered by syikin_310 2 · 0 1

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