# Solving $\log _2(x-4) + \log _2(x+2) = 4$

Here is how I have worked it out so far:

$\log _2(x-4)+\log(x+2)=4$

$\log _2((x-4)(x+2)) = 4$

$(x-4)(x+2)=2^4$

$(x-4)(x+2)=16$

How do I proceed from here?

$x^2+2x-8 = 16$

$x^2+2x = 24$

$x(x+2) = 24$ Which I know is not the right answer

$x^2+2x-24 = 0$ Can't factor this

• Yes it makes it alot easier to help when you show the work like this. Nov 18, 2010 at 0:55
• Something like this in mathematica simply checks for error :Solve[Log[2, x - 4] + Log[2, x + 2] == 4, x] Nov 18, 2010 at 9:58

It is $x^2-2x-8 = 16$ my friend. So you get $x^2 - 2x -24 = 0$, which factors as $(x-6)(x+4) = 0$. Hence, $x=6$ or $x = -4$.

• I suspect that the x=-4 root is meant to be discarded (otherwise your claiming log_2(-8)+log_2(-2)=4). Nov 18, 2010 at 0:49
• @ Douglas S. Stones: which is not totally false since I could take a branch of $\log_2(-8)$ and another branch of $\log_2(-2)$ to get $4$. But yes if you are talking about "conventional" logarithms, I need to discard $x=-4$.
– user17762
Nov 18, 2010 at 7:29

After $(x-4)(x+2)=16$, you get $x^2-2x-24=0$ (the coefficient of $x$ is $-2$ not $2$). So $x=\frac{2\pm \sqrt{100}}{2}$ by the quadratic formula. So $x=6$ or $x=-4$

• In the square root, how do you get 100? 4 - (4)(1)(24) = 4 - (96) = 92 Nov 18, 2010 at 0:24
• $4 - (4)(1)(-24)=100$. The constant is $-24$, not $24$ Nov 18, 2010 at 0:27
• Ah, I get it. Negative numbers are going to be the end of my math career :P Nov 18, 2010 at 0:32

The quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4 a c} }{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24) }}{2 \times 1}$$ 4 + 96 in the square root.