Solve for x. #log_x (log_3 x) = 2#?

Using the formula

#log_b x = lnx/lnb#

we can rewrite our equation as

#ln(lnx/ln3)/lnx=2#

#ln(lnx/ln3) = ln(x^2)#

#ln(lnx/x^2 * 1/ln3) = 0#

We know that #ln 1 = 0#, hence

#lnx/x^2 *1/ln3 = 0 = > lnx/x^2=ln3#

#lnx = x^2ln3#

When we have an equation of this type, we want the arguement of the natural logarithm to have the same exponent as #x# in the right hand side (2, in this case). To do this, we must multiply both sides by two and use the property that:

#clog_b a = log_b a^c#

#:.ln(x^2)=2x^2ln3#

Raise #e# to both sides:

#x^2=e^(x^2 * 2ln3)#

Since #2ln3 = ln9#, we have

#x^2=9^(x^2)#

Let's solve a more general case. Consider the equation

#z=C^z#

for #z in CC#.

#z/C^z = 1#

#ze^(-zlnC)=1# #color(white)(ffffff) (E_1)#

The thought process here is that we wish to have everything in terms of #e# and the natural logarithm; this way, we can apply use the Lambert W function, which is defined as the inverse of the function

#f(w) = we^w#

#color(red)(W = f^(-1))#

As such, in the equation

#we^w = a#

we see that:

#a=f(w) => w=f^(-1)(a) = W(a)#

Now, multiply both sides of #E_1# by #-lnC#:

#-zlnCe^(-zlnC) = -lnC#

Let #z_0 = -zlnC#.

#z_0e^(z_0) = - lnC#

#color(blue)( => z_0 = W(-lnC))#

Finally, we have reached a solution! Undoing all substitutions, we get:

#-zlnC=W(-lnC) => color(blue)(z = (W(-lnC))/(-lnC))#

We have proven that the solution of the equation

#z=C^z# is

#z=(W(-lnC))/(-lnC)#

Our equation looked like this

#x^2=9^(x^2)#

#:. x^2=(W(-ln9))/(-ln9) => color(red)(x=sqrt((W(-ln9))/(-ln9)))#

This is the exact form of #x#. As it turns out, #x# is complex; its numerical value is approximately

#x ~~ 0.581 +- 0.669 i#