#5e^x = x^3#, find #x#?
1 Answer
where
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
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
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
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.