# Solution… 2nd Transcendental Equation with Graphical method [closed]

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

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

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• Τhanks Αλέξανδρος for formatting!!! – Nikos Mantzakouras Dec 9 '18 at 6:47

## 3 Answers

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!

• Yes ,,, you are right, it is simply called and so...Τhanks ..Sure it has and complex Roots!!! – Nikos Mantzakouras Dec 8 '18 at 18:06
• 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[]]} – Nikos Mantzakouras Dec 8 '18 at 18:23
• The complex roots is ONLY 3 because for any case 0<=k’<=2!!! – Nikos Mantzakouras Dec 8 '18 at 18:44
• W-function is ProductLog() – Nikos Mantzakouras Dec 8 '18 at 18:47

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}}

• Yes it is correct!!! – Nikos Mantzakouras Dec 8 '18 at 18:34

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] {{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}]}}

• You dont know Abs[x]<2,,FROM WHERE? – Nikos Mantzakouras Dec 10 '18 at 9:58