How do you solve ${log}_{3} (5 x + 5) - {log}_{3} (x^{2} - 1) = 0$ $log_3 (5x+5) - log_3 (x^2 - 1) = 0$ ?

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Answer 1 · 2015-05-11T23:14:32

The answer is $x = 6$ $x=6$

I recall the rule:
$log (a) - log (b) = log (\frac{a}{b})$ $log(a)-log(b)=log(a/b)$
and $log (a) = 0 \Leftrightarrow a = 1$ $log(a)=0 <=> a=1$

so ${log}_{3} (\frac{5 x + 5}{x^{2} - 1}) = 0 \Leftrightarrow$ $log_3((5x+5)/(x^2-1))=0 <=>$

$\Leftrightarrow \frac{5 x + 5}{x^{2} - 1} = 1 \Leftrightarrow 5 x + 5 = x^{2} - 1$ $<=> (5x+5)/(x^2-1)=1 <=> 5x+5=x^2-1$

So we gotta solve $x^{2} - 5 x - 6 = 0$ $x^2-5x-6=0$

$\frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm 7}{2} \Rightarrow x = 6 or x = - 1$ $(5+-sqrt(25+24))/2 = (5+-7)/2 => x=6 or x=-1$

6 is an acceptable solution:

$5 \cdot 6 + 5 = 35 > 0$ $5*6+5=35>0$ and $36 - 1 = 35 > 0$ $36-1=35>0$

-1 is not an acceptable answer:
$- 5 + 5 = 0 and log (0)$ $-5+5=0 and log(0)$ is not defined, so that solution is an extraneous solution.

How do you solve log3(5x+5)−log3(x2−1)=0log_3 (5x+5) - log_3 (x^2 - 1) = 0?