Im missing something. Can anyone help?

log_2 (x) + log_2 (x-4) = 2

=> log_2 (x^2-4x) = 2

=> x^2-4x = 2^2

=> where do I go from here? Ive tried various methods and cant come up with the right answer. Everything is right up to this point right?

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- Mar 27th 2009, 10:21 AM #1

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- Mar 27th 2009, 10:24 AM #2

- Mar 27th 2009, 10:27 AM #3

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- Mar 27th 2009, 10:35 AM #4
I used the quadratic formula. For $\displaystyle ax^2 +bx+c = 0$:

$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where a = 1, b=-4 and c=-4

Sub them in:

$\displaystyle x = \frac{4 \pm \sqrt{4^2-4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2}$

as $\displaystyle \sqrt{32} = \sqrt{16}\sqrt{2} = 4\sqrt{2}$ we can simplify and factorise:

$\displaystyle \frac{4(1 \pm \sqrt{2})}{2} = 2(1 \pm \sqrt{2})$.

Since $\displaystyle 2(1-\sqrt{2}) < 4$ we discard it as a solution because $\displaystyle log_2(x-4)$ is undefined.

$\displaystyle 2(1+\sqrt{2}) > 4$ so this is a solution. You can either leave it in surd form for an exact answer or round it off

- Mar 27th 2009, 11:32 AM #5

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