Solution Transcedental Equation with(Graphical method)

Question

If 3x + logx = 30, then what is x?

Solution with W-Function... The equation is 3x+Logx=30(1) we know that accept we^w=y=>w=W(k,y)..k in Z.But if w=3x=>3xe^(3x)=y=>3x=W(k,y)(2) but from (1)… Log(3x)+3x=Logy=>Log3+Logx+3x=Logy=> Logx+3x=Logy-Log3=Log(y/3). Because we have the same equation {1,2} and then Log(y/3)=30=>y=3*e^30.Then General Solution is

3x=W(k,y)=>x=1/3W(k,3*e^30),, k in Z.

Program in mathematica .....................
Reduce[3*x + Log[x] == 30, x],,
C[1] [Element] Integers && x == 1/3 ProductLog[C[1], 3 E^30] — Nikos Mantzakouras, Nov 15 '18 at 20:36
in() ..Solve[3*x + Log[x] == 30, x, Reals] ..out() {{x -> 1/3 ProductLog[3 E^30]}}..This mean C(1)=0 from before.. — Nikos Mantzakouras, Nov 15 '18 at 20:46
Does Plot[{Log[x], 30 - 3 x}, {x, 1, 12}] count as a graphical method of solving this? — Bill, Nov 15 '18 at 23:11
Possible duplicate: https://mathematica.stackexchange.com/questions/10472/marking-points-of-intersection-between-two-curves — Michael E2, Nov 16 '18 at 02:16

score 1 · Answer 1 · answered Nov 16 '18 at 02:35

For a purely graphical approach, you can use the options MeshFunctions and Mesh and post-process the Plot output to add text elements:

f[x_] = 3 x + Log[x] - 30;
Normal[Plot[f[x], {x, 0, 20}, 
   PlotPoints -> 100,
   MeshFunctions -> {#2 &}, 
   Mesh -> {{0}}, 
   MeshStyle -> Directive[Red, PointSize[Large]]]] /. 
  p_Point :> {p, Text[Style[p[[1, 1]], 16], {-1, 5} + p[[1]]]}

score 0 · Answer 2 · answered Nov 15 '18 at 23:07

0

Normally transcendental equations are numerically solved with NSolve or FindRoot.

f[x_] = 3 x + Log[x] - 30;

Plot[f[x], {x, 0, 20}]

NSolve[f[x] == 0 && x > 5 && x < 15]
(* {{x -> 9.25816}} *)

FindRoot[f[x] == 0, {x, 10}]
(* {x -> 9.25816} *)

answered Nov 15 '18 at 23:07

Bill Watts

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ok many thanks is correct – Nikos Mantzakouras Nov 16 '18 at 11:44

Solution Transcedental Equation with(Graphical method)

2 Answers2