How do I solve for x ?

#log_ax=x^2/a^2#

1 Answer
Feb 26, 2018

Apart from #x=a# the other solution can be found using the Lambert W function.

Explanation:

Given:

#log_a x = x^2/a^2#

Note that one straightforward solution is #x = a# since:

#log_a a = 1 = a^2/a^2#

Beyond that simple solution, we need to use the Lambert W function...

We want to get this into a form:

#z e^z = w#

so we can use the Lambert W function.

Using the change of base formula, we can write:

#x^2/a^2 = ln x / ln a = ln x^2 / (2 ln a)#

So:

#(2 ln a)/a^2 x^2 = ln x^2#

Taking the exponent of both sides:

#e^((2 ln a)/a^2 x^2) = x^2#

Multiplying both sides by #e^(-(2 ln a)/a^2 x^2)#:

#1 = x^2 e^(-(2 ln a)/a^2 x^2)#

Multiplying both sides by #-(2 ln a)/a^2#:

#-(2 ln a)/a^2 = (-(2 ln a)/a^2 x^2) e^(-(2 ln a)/a^2 x^2)#

Applying the Lambert W function to both sides:

#W(-(2 ln a)/a^2) = -(2 ln a)/a^2 x^2#

Multiplying both sides by #-a^2/(2 ln a)# this becomes:

#x^2 = -a^2/(2 ln a) W(-(2 ln a)/a^2)#

Then taking the positive square root of both sides, we get:

#x = sqrt(-a^2/(2 ln a) W(-(2 ln a)/a^2))#