How do you solve the equation ${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2(12b-21)-log_2(b^2-3)=2$ ?

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How do you solve the equation ${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2(12b-21)-log_2(b^2-3)=2$ ?

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Answer 1 · 2017-01-10T17:42:10

We do not have a solution to ${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2 (12b-21)-log_2 (b^2-3)=2$

Explanation:

${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2 (12b-21)-log_2 (b^2-3)=2$

$\Leftrightarrow {log}_{2} (\frac{12 b - 21}{b^{2} - 3}) = 2$ $hArrlog_2 ((12b-21)/(b^2-3))=2$

or ${log}_{2} (\frac{12 b - 21}{b^{2} - 3}) = {log}_{2} 2^{2}$ $log_2 ((12b-21)/(b^2-3))=log_2 2^2$

or $\frac{12 b - 21}{b^{2} - 3} = 4$ $(12b-21)/(b^2-3)=4$

or $12 b - 21 = 4 b^{2} - 12$ $12b-21=4b^2-12$

or $4 b^{2} - 12 - 12 b + 21 = 0$ $4b^2-12-12b+21=0$

or $4 b^{2} - 12 b + 9 = 0$ $4b^2-12b+9=0$

or ${(2 b)}^{2} - 2 \times 2 b \times 3 + 3^{2} = 0$ $(2b)^2-2xx2bxx3+3^2=0$

or ${(2 b - 3)}^{2} = 0$ $(2b-3)^2=0$

i.e. $2 b - 3 = 0$ $2b-3=0$ or $b = \frac{3}{2}$ $b=3/2$

However, if $b = \frac{3}{2}$ $b=3/2$ , ${log}_{2} (12 b - 21)$ $log_2 (12b-21)$ or ${log}_{2} (b^{2} - 3)$ $log_2 (b^2-3)$ do not exist as $12 b - 21$ $12b-21$ and $b^{2} - 3$ $b^2-3$ are negative.

Hence, we do not have a solution to ${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2 (12b-21)-log_2 (b^2-3)=2$

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How do you solve the equation ${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2(12b-21)-log_2(b^2-3)=2$ ?

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Explanation:

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How do you solve the equation log2(12b−21)−log2(b2−3)=2log_2(12b-21)-log_2(b^2-3)=2?

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Explanation:

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How do you solve the equation ${log}_{2} (12 b - 21) - {log}_{2} (b^{2} - 3) = 2$ $log_2(12b-21)-log_2(b^2-3)=2$ ?