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If $ 2x^3 + \ln x = 5 $, then what is $ x $?

For beginning we started to find a solution for equation

$$ \begin{align}\label{eq:eq} \ln x + cx = b \tag{1} \end{align} $$

We know from W function that $ ax\mathrm e^{ax} = y \Rightarrow ax = W(k, y) $, where $ k \in \mathbb Z $. From Eq. \ref{eq:eq} $$ \begin{align} \Rightarrow \ln(ax) + ax &= \ln y \\ \Rightarrow \ln x + x &= \ln(y/a) \label{eq:eq2}\tag{2} \end{align} $$

From Eqs. \ref{eq:eq} and \ref{eq:eq2}

=>a=c & Log(y/a)=b.Then y=ae^(b) & x=1/aW(k,a*e^(b))..k in Z.(3).If now we take the original equation Log x+2*x^3=5 we do the transformation …x^3=z (5) =>3*Log(x)=Log(z)+2kπi =>x=z^(1/3)*e^(2k’πi/3)..k’ in Z (6) Βut with the transformation(5) the relation (1) is done Log(z)+6z=15.. (7).But the relation (7) has solution in accordance with the foregoing z=1/6*W(k,6e^(15))..(8) ,k in Z.From (6&8) we have the filnal solution =>

x=(1/6*W(k,6e^(15)))^(1/3)*e^(2k’πi/3) ,k&k’ in Z. The solutions are 3 only …

1…with k=0 & k’=0,…,x=1.33084 ,,,Real and we have 2 complex roots 2.. for k=1 & k’=2 =>x=-0.520715 - 1.26144 I and 3.. for k=-1 & k’=-2 =>x=-0.520715 + 1.26144 I

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closed as off-topic by corey979, Michael E2, eyorble, Bob Hanlon, Αλέξανδρος Ζεγγ Dec 9 '18 at 3:04

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  • $\begingroup$ Τhanks Αλέξανδρος for formatting!!! $\endgroup$ – Nikos Mantzakouras Dec 9 '18 at 6:47
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What about

Solve[2 x^3 + Log[x] == 5, x] 
(*{{x -> (1/6 ProductLog[6 E^15])^(1/3)}}*)

What you cal "W-function" is Mathematica ProductLog!

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  • $\begingroup$ Yes ,,, you are right, it is simply called and so...Τhanks ..Sure it has and complex Roots!!! $\endgroup$ – Nikos Mantzakouras Dec 8 '18 at 18:06
  • $\begingroup$ Graphical Method........in Mathematica!!! f[x_] = 2 x^3 + Log[x] - 5; Normal[Plot[f[x], {x, 0, 2}, PlotPoints -> 100, MeshFunctions -> {#2 &}, Mesh -> {{0}}, MeshStyle -> Directive[Red, PointSize[Large]]]] /. p_Point :> {p, Text[Style[p[[1, 1]], 14], {0, 5} + p[[1]]]} $\endgroup$ – Nikos Mantzakouras Dec 8 '18 at 18:23
  • $\begingroup$ The complex roots is ONLY 3 because for any case 0<=k’<=2!!! $\endgroup$ – Nikos Mantzakouras Dec 8 '18 at 18:44
  • $\begingroup$ W-function is ProductLog() $\endgroup$ – Nikos Mantzakouras Dec 8 '18 at 18:47
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FindInstance can locate all three roots:

FindInstance[2 x^3 + Log[x] == 5, x, 3] // N

{{x -> -0.520715 - 1.26144 I}, {x -> 1.33084}, {x -> -0.520715 + 1.26144 I}}

Alternatively, FindRoot can find all three, though you have to help it with different starting points:

FindRoot[2 x^3 + Log[x] == 5, {x, #}] & /@ {1, -1 + I, -1 - I}

{{x -> 1.33084}, {x -> -0.520715 + 1.26144 I}, {x -> -0.520715 - 1.26144 I}}
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  • $\begingroup$ Yes it is correct!!! $\endgroup$ – Nikos Mantzakouras Dec 8 '18 at 18:34
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Solve or NSolve will give all three roots if you constrain the Abs value of x

Solve[{2 x^3 + Log[x] == 5, Abs[x] < 2}, x]

enter image description here

{{x -> Root[{-5 + Log[#1] + 
       2 #1^3 &, \
-0.52071466111311432025020844963466209285486182967411316474958 - 
      1.26143852926442301838345194738901835961144796731648927183682 I}]}, {x \
-> Root[{-5 + Log[#1] + 
       2 #1^3 &, \
-0.52071466111311432025020844963466209285486182967411316474958 + 
      1.26143852926442301838345194738901835961144796731648927183682 I}]}, {x \
-> Root[{-5 + Log[#1] + 2 #1^3 &, 1.33083954213436297147435824439}]}}
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  • $\begingroup$ You dont know Abs[x]<2,,FROM WHERE? $\endgroup$ – Nikos Mantzakouras Dec 10 '18 at 9:58

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