How do you solve #log_[2](12b-21) - log_[2](b^2-3) = 2#?

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Answer 1 · 2016-06-17T01:12:15

Use the rule #log_a(n) - log_a(m) = log_a(n/m)#

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

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

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

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

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

#0 = 4b^2 - 6b - 6b + 9#

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

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

#b = 3/2#

However, checking in the original equation, we find this solution doesn't work.

Therefore, the solution set is #{x = O/}#.

Hopefully this helps!