#5e^x = x^3#, find #x#?

1 Answer
Jul 16, 2018

#x = -3 W_n(-5^(1/3)/3 omega^m)" "# for any #n in ZZ# and #m in { 0, 1, 2 }#

where #W_n# is the Lambert W function and #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#

Explanation:

This kind of equation has no solution expressible in terms of elementary functions, but we can use a family of functions called the Lambert #W# function to help.

An equation of the form:

#z e^z = c#

has solutions:

#z = W_n(c)" "# for #n in ZZ#

Given:

#5e^x = x^3#

Taking into account all cube roots, this means:

#5^(1/3) omega^m e^(x/3) = x" "# for #m in { 0, 1, 2 }#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

Then:

#5^(1/3)/3 omega^m e^(x/3) = x/3#

Then:

#-5^(1/3)/3 omega^m = (-x/3) e^((-x/3))#

Note that the right hand side is in the form #z e^z# with #z = -x/3#.

Hence we can apply the Lambert W function to find:

#-x/3 = W_n(-5^(1/3)/3 omega^m)" "# for any #n in ZZ# and #m in { 0, 1, 2 }#

So:

#x = -3 W_n(-5^(1/3)/3 omega^m)" "# for any #n in ZZ# and #m in { 0, 1, 2 }#

This only takes non-real complex values.