Question
Solve the following:$\log _8 (x^2- 1) - \log _8 (3x + 9) = 0$
✓
Answer
$\log _8\left(x^2-1\right)-\log _8(3 x+9)=0$
$\Rightarrow \log _8\left(\frac{x^2-1}{3 x+9}\right)=\log _8 1$
$\Rightarrow \frac{x^2-1}{3 x+9}=1$
$\Rightarrow x^2-1=3 x+9$
$\Rightarrow x^2-3 x-10=0$
$\Rightarrow x^2-5 x+2 x-10=0$
$\Rightarrow x(x-5)+2(x-5)=0$
$\Rightarrow(x-5)(x+2)=0$
$\Rightarrow x=5 \text { or } x=-2$
$\Rightarrow \log _8\left(\frac{x^2-1}{3 x+9}\right)=\log _8 1$
$\Rightarrow \frac{x^2-1}{3 x+9}=1$
$\Rightarrow x^2-1=3 x+9$
$\Rightarrow x^2-3 x-10=0$
$\Rightarrow x^2-5 x+2 x-10=0$
$\Rightarrow x(x-5)+2(x-5)=0$
$\Rightarrow(x-5)(x+2)=0$
$\Rightarrow x=5 \text { or } x=-2$
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