How do you solve the equation #log_2(12b-21)-log_2(b^2-3)=2#?

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Answer 1 · 2017-01-10T17:42:10

We do not have a solution to #log_2 (12b-21)-log_2 (b^2-3)=2#

#log_2 (12b-21)-log_2 (b^2-3)=2#

#hArrlog_2 ((12b-21)/(b^2-3))=2#

or #log_2 ((12b-21)/(b^2-3))=log_2 2^2#

or #(12b-21)/(b^2-3)=4#

or #12b-21=4b^2-12#

or #4b^2-12-12b+21=0#

or #4b^2-12b+9=0#

or #(2b)^2-2xx2bxx3+3^2=0#

or #(2b-3)^2=0#

i.e. #2b-3=0# or #b=3/2#

However, if #b=3/2#, #log_2 (12b-21)# or #log_2 (b^2-3)# do not exist as #12b-21# and #b^2-3# are negative.

Hence, we do not have a solution to #log_2 (12b-21)-log_2 (b^2-3)=2#