How do you solve #log_x 2 + log_2 x = 3# ?

First we will convert #log_x(2)# to #log_2#. We will use the following log property:
#log_b(a)=log_x(a)/log_x(b)#

In our equation, this becomes:
#log_x(2)+log_2(x)=log_2(2)/log_2(x)+log_2(x)=1/log_2(x)+log_2(x)#

This is a quadratic, but to make it easier to understand I will introduce a substitution so #u=log_2(x)#:
#1/log_2(x)+log_2(x)=3#

#1/u+u=3#

Multiply by #u# on both sides:
#1+u^2=3u#

#u^2-3u+1=0#

Solve using the quadratic formula:
#u=(3+-sqrt5)/2#

Now we resubstitute, so we have
#log_2(x)=(3+-sqrt5)/2#

We put both sides as powers of #2#:
#cancel(2)^(cancel(log_2)(x))=2^((3+-sqrt5)/2)#

#x=2^((3+-sqrt5)/2)#

Given:

#log_x 2 + log_2 x = 3#

By the change of base formula, we have:

#log 2 / log x + log x / log 2 = 3#

Letting #t = log x / log 2 = log_2 x#, that becomes:

#1/t + t = 3#

Mutliplying through by #t# and rearranging a bit:

#0 = 4(t^2-3t+1)#

#color(white)(0) = 4t^2-12t+4#

#color(white)(0) = (2t)^2-2(2t)(3)+3^2-5#

#color(white)(0) = (2t-3)^2-(sqrt(5))^2#

#color(white)(0) = ((2t-3)-sqrt(5))((2t-3)+sqrt(5))#

#color(white)(0) = (2t-3-sqrt(5))(2t-3+sqrt(5))#

So:

#2t = 3+-sqrt(5)#

That is:

#2log_2 x = 3+-sqrt(5)#

So:

#x = 2^(1/2(3+-sqrt(5))#