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#
