How do you solve ${log}_{2} (9 x + 5) = 2 + {log}_{2} (x^{2} - 1)$ $log_2(9x+5)=2+log_2(x^2-1)$ ?

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How do you solve ${log}_{2} (9 x + 5) = 2 + {log}_{2} (x^{2} - 1)$ $log_2(9x+5)=2+log_2(x^2-1)$ ?

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Answer 1 · 2016-10-16T16:00:30

$x = - \frac{3}{4} or 3$ $x=-3/4 or3$

Explanation:

We need to write 2 in the form of ${log}_{2} ?$ $log_2?$
$2^{2} = 4$ $2^2=4$ so ${log}_{2} 4 = 2$ $log_2 4=2$

Rewrite the equation as
${log}_{2} (9 x + 5) = {log}_{2} 4 + {log}_{2} (x^{2} - 1)$ $log_2(9x+5)=log_2 4+log_2(x^2-1)$
Or using the addition rule of logs
${log}_{2} (9 x + 5) = {log}_{2} 4 (x^{2} - 1)$ $log_2(9x+5)=log_2 4(x^2-1)$
Therefore
$9 x + 5 = 4 (x^{2} - 1)$ $9x+5=4(x^2-1)$
Rearrange
$9 x + 5 = 4 x^{2} - 4$ $9x+5=4x^2-4$
$4 x^{2} - 9 x - 9 = 0$ $4x^2-9x-9=0$
Factorise
$(4 x + 3) (x - 3) = 0$ $(4x+3)(x-3)=0$
Thus $x = - \frac{3}{4} or 3$ $x=-3/4 or3$

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How do you solve ${log}_{2} (9 x + 5) = 2 + {log}_{2} (x^{2} - 1)$ $log_2(9x+5)=2+log_2(x^2-1)$ ?

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Explanation:

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How do you solve ${log}_{2} (9 x + 5) = 2 + {log}_{2} (x^{2} - 1)$ $log_2(9x+5)=2+log_2(x^2-1)$ ?