log base 2 of (x^2-4x) < 5

solve the inequality. give answers in interval form.

2. Hi Mr. Green,

Given $\log_2 (x^2-4x) < 5$.

Since exponential is a monotonic increasing function, we can raise both sides to the power of 2 and preserve the inequality. Get:

$x^2-4x < 32$.
$x^2-4x-32 < 0$.

Factor:

$(x-8)(x+4)<0$.

Can you see the interval of valid x's from here?
Hint: try plugging in a few x's and seeing which ones are true for the inequality and which ones aren't, and construct your interval from that.

3. ## rite?

i got

(-inf, -4) U (8, +pi)

is it rite?

4. Originally Posted by Mr_Green
i got

(-inf, -4) U (8, +pi)

is it rite?
No. First of all $(8,\pi)$ doesn't make sense as an interval because $8 > \pi$.

However, to ask yourself if you're right, try one of the values.
Try -10, which is in your first interval there.

$((-10)-8)((-10)+4) = (-18)(-6) = 108$.
108 is certainly not less than zero, so right there that should set off alarm bells.

Always a good number to try in a situation like this is 0.
$((0)-8)((0)+4) = (-8)(4) = -32$.
Certainly -32 < 0, so we know 0 lies in the interval we're seeking!

Now test some numbers around zero and see where the interval breaks off. You know somewhere it stops between -10 and 0, since -10 is not in the interval and 0 is.

Look at the final formula itself. What information can you pry out of it?

5. i meant +inf and not pi. my fault...??

6. Originally Posted by Mr_Green
i meant +inf and not pi. my fault...??
Regardless, even if it were infinity, your interval is wrong. In fact, it is very close to exactly the opposite.

The interval you're looking for is this: $(-4,8)$.
Try to understand why though, and notice how similar those numbers are to numbers found in the final factored form.