Page 32 - Calculus Workbook For Dummies

16 Part I: Pre-Calculus Review

$
n Rewrite log5 + log200 with a single log and then solve. log5 + log200 = log 5 200 = log1000 = 3
h
^
When you see “log” without a base number, the base is 10.

8
o If x5 2 = 3 x + 8, solve for x with the quadratic formula. x = or - 1
5
2
5
8
Start by rearranging x5 2 = 3 x + 8 into x - 3 x - = 0 because you want just a zero on one side
of the equation.
2
- b ! b - 4 ac
The quadratic formula tells you that x = . Plugging 5 into a, –3 into b, and –8
2 a
2
- - 3 ! ^h - 3 - ^h 4 5 - 8h 3 ! 9 + 160 3 ! 13 16 - 10
^
^ h
into c gives you x = = = = or ,
2 5 $ 10 10 10 10
8
so x = or - 1.
5
16
x
p Solve x3 + 2 > 14. < - , x > 4
3
1. Turn the inequality into an equation: x3 + 2 = 14
2. Solve the absolute value equation.
2
3 x + = 14 3 x + = - 14
2
3 x = 12 or 3 x = - 16
x = 4 16
x = -
3
3. Place both solutions on a number line (see the following figure). (You use hollow dots for
> and <; if the problem had been $ or #, you would use solid dots.)
-16 4
3
4. Test a number from each of the three regions on the line in the original inequality.
For this problem you can use –10, 0, and 10.
?
3 $ ^ - 10 + 2 >14
h
?
- 28 >14
?
28 >14
True, so you shade the left-most region.
?
3 $ ^ h 2 >14
0 +
?
2 >14
False, so you don’t shade the middle region.
?
$
3 10 + 2 >14
?
32 >14
?
32 >14
True, so shade the region on the right. The following figure shows the result. x can be any
number where the line is shaded. That’s your final answer.

-16 4
3