Question
Solve the following:$\log _4 x+\log _4(x-6)=2$
✓
Answer
$\log _4 x + \log_4(x-6) = 2$
$\Rightarrow \log _4{x(x-6)} = 2 \log _4 4$
$\Rightarrow \log _4 {x^2- 6x} = \log _4 4^2$
$\Rightarrow x^2 - 6x = 16$
$\Rightarrow x^2 - 6x - 16 = 0$
$\Rightarrow x^2 - 8x + 2x - 16 = 0$
$\Rightarrow x (x - 8) + 2 ( x - 8) = 0$
$\Rightarrow (x - 8)(x + 2) = 0$
$\Rightarrow x = 8$ or $-2$
Negative value is rejected
So, $x = 8$.
$\Rightarrow \log _4{x(x-6)} = 2 \log _4 4$
$\Rightarrow \log _4 {x^2- 6x} = \log _4 4^2$
$\Rightarrow x^2 - 6x = 16$
$\Rightarrow x^2 - 6x - 16 = 0$
$\Rightarrow x^2 - 8x + 2x - 16 = 0$
$\Rightarrow x (x - 8) + 2 ( x - 8) = 0$
$\Rightarrow (x - 8)(x + 2) = 0$
$\Rightarrow x = 8$ or $-2$
Negative value is rejected
So, $x = 8$.
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