1. ## Log help

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?

2. Originally Posted by Jimbo
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?
You're right up to now yes. Take 4 from both sides:

x^2 - 4x - 4 = 0 and solve using your favourite method. Remember that x > 4 because you can't take logs of 0 or negative numbers

3. Thats where Im running into trouble. Ive tried several methods to solve for x and cant seem to come up right. Ive got to go for now, but Ill come back later and post some things ive tried. Thanks!

4. Originally Posted by Jimbo
Thats where Im running into trouble. Ive tried several methods to solve for x and cant seem to come up right. Ive got to go for now, but Ill come back later and post some things ive tried. Thanks!
I used the quadratic formula. For $ax^2 +bx+c = 0$:

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

Sub them in:

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

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

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

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

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

5. Right after I logged off I remembered the quadratic formula and was going to try it out. I always forget about that.

Thanks for the help!