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

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