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Question Number 12279 by chux last updated on 17/Apr/17
i guess i posted this question but  didnt really get much response.  pls help out    if 8log_2  x + (x−1)log_x  2 =3^x    find x.
iguessipostedthisquestionbutdidntreallygetmuchresponse.plshelpoutif8log2x+(x1)logx2=3xfindx.
Commented by mrW1 last updated on 17/Apr/17
there is no analytical solution for  such equation. by trial and error you  can see that x=2 is a solution. using  graphical method you can find the  other solution x≈1.3853.
thereisnoanalyticalsolutionforsuchequation.bytrialanderroryoucanseethatx=2isasolution.usinggraphicalmethodyoucanfindtheothersolutionx1.3853.
Commented by mrW1 last updated on 18/Apr/17
the newton raphson iteration to solve  f(x)=0 is:  if you have a start value x_0  for the solution,  you can get a better value with  x_(n+1) =x_n −((f(x_n ))/(f′(x_n )))     (n=1,2,3....)  you repeat this till x_(n+1) −x_n  is small  enough for you.    your case is not good for demonstration.  let us see f(x)=sin x−0.4=0  f′(x)=cos x  x_0 =0.4  x_1 =0.4−((sin 0.4−0.4)/(cos 0.4))=0.411488553  x_2 =0.411488553−((sin 0.411488553−0.4)/(cos 0.411488553))=0.411516846  x_3 =....=0.411516846≈x_2   we get after 3 iterations a good solution  for sin x−0.4=0:   x≈0.411516846
thenewtonraphsoniterationtosolvef(x)=0is:ifyouhaveastartvaluex0forthesolution,youcangetabettervaluewithxn+1=xnf(xn)f(xn)(n=1,2,3.)yourepeatthistillxn+1xnissmallenoughforyou.yourcaseisnotgoodfordemonstration.letusseef(x)=sinx0.4=0f(x)=cosxx0=0.4x1=0.4sin0.40.4cos0.4=0.411488553x2=0.411488553sin0.4114885530.4cos0.411488553=0.411516846x3=.=0.411516846x2wegetafter3iterationsagoodsolutionforsinx0.4=0:x0.411516846
Commented by chux last updated on 18/Apr/17
thanks sir..... a friend told me about  a method called newton raphson  iteration but i dnt knw how it can  be applied.... please can you   explain the method and applicatn  to thiz question if possible....      thanx for always helping.
thankssir..afriendtoldmeaboutamethodcallednewtonraphsoniterationbutidntknwhowitcanbeapplied.pleasecanyouexplainthemethodandapplicatntothizquestionifpossible.thanxforalwayshelping.
Commented by chux last updated on 18/Apr/17
thanks boss
thanksboss

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