This suggests that you can use #log_12(x)#, where #x# is either #12^(2x-3)# or #16#. (By default, #log x = log_10 x#.)

#log_12(12^(2x-3)) = log_12(16)#

Now if you use the change of base law, you get:

#(log (12^(2x-3)))/(log12) = (log 16)/(log 12)#

Since #(lnx)/(logx) = "constant"# (try it on any #x#; it is #~~2.303#), we can change this to:

#(ln (12^(2x-3)))/(ln12) = (ln 16)/(ln 12)#

Using exponential rules of logs:

#(2x-3)cancel(((ln 12)/(ln12))) = (ln 16)/(ln 12)#

#2x - 3 = (ln 16)/(ln 12)#

#x = color(blue)(((ln 16)/(ln 12) + 3)/2 ~~ 2.05789) #

If you plug this back in, it indeed works.

#12^((2(~~2.05789) - 3)) = 16#

#12^((~~1.11578)) = 16#