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1
Solve for x. 4(3x -2) =
256. (See example 1 a. ) The value of x is in the ranges below:
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a (0,3)
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| b (3,6) |
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c (6,9)
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d (9,12)
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e nota
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2. 125x = 625 ( See example 1 b. Hint write 625 as a power of 5; Also
write 125 as a power of 5. On the 125x expression you will get
125 = 5nx, where n is an integer. Then solve for x like
you did in the previous problem, yo. x is in the range:
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a (0,2)
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b (2,4)
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c (4,6)
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d nota
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3. Solve for x . 9ex = 99. Hint : first
divide by 9 to get ex = 11. Then use the golden rule for logs
as used in quiz 12. Or you can take the natural log (ln = loge)
of both sides. In any event, see examples 3 and 8.
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| a x = log9 11
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b x = log 11 |
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c x = ln 11 |
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d x = log 99 |
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e nota
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4. Solve for x. log3x = 4. See quiz 12 and example 4
of this section (9.5). |
| a x = 3 |
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b x = 27 |
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c x = 81 |
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d nota
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| 5. Solve for x. log5(x
- 7) = 2. See example 4. |
| a x = 12 |
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b x = 22 |
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c x = 32 |
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d x = 42 |
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e nota
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| 6. Rewrite the following in
exponential form: log6(x + 5) + log6x
= 2. Example 5.
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| a ( x + 5) + x = 36 |
| b ( x + 5)·x = 36 |
| c ( x + 5)·x = 12 |
| d ( x + 5)·x = 8 |
| nota
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| 7. Solve for x in the previous
problem. x is in the following ranges:
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a (-10, 0) or ( 0, 5) |
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b (-5, 0) or (0,10) |
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c (0,5) or (5,10) |
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d nota |
8. Example 5. Solve for x: log3(x
- 5) + log3(x + 3)
= 2.
Hint: Rewrite as log3(x - 5)·(x
+ 3)
= 2. Then write in exponential form. The value for x is in the ranges: |
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a (0,3) or (3,7) |
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b (-7,0) or (0,3) |
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c (-7,0) or (0,7) |
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d nota |
| written problems: |
9. See example 6. Solve for x. log(2x -
1) - logx
= 2. Yo, log means log base 10 and subtraction must be converted to log
of the quotient. Once you do that, convert to exponential notation, then
cross multiply. Hint: Rewrite the equation as:
log[(2x -
1)/x] = 2
log10[(2x -
1)/x]
= 2
Now write in log form:
102 = (2x -
1)/x
100 = (2x -
1)/x
Now cross multiply and solve for x.
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| 10.1 |
| 10. Find the distance d between the
points (-6,-1) and (2,5). Answer: distance d = 10 (See
example 1) |
| 11. Find the midpoint between the
points (-6,-1) and (2,5). Answer: Midpoint = (-2, 2) (See
example 2) |
12. Find the center and radius
of the circle given by the equation here.
(x - 1)2 + (y + 2)2 = 256. Answer: (center =
(1, -2) and radius = 16. Note 162 = 256. ( See example 3) |
13. Find the center and radius of the circle given by the
equation here. ( See example 6)
x2 + y2 + 6x + 2y + 6 = 0.
Hint:
# Rewrite the equation as x2 + 6x + y2
+ 2y + 6 = 0.
#The complete the square twice for the terms that are in brackets:
(x2 + 6x) + (y2 + 2y) + 6 = 0
(x2 + 6x + 9) + (y2 + 2y + 1) + 6
= 9 + 1. Note in this step, I had to add 9 + 1 to both sides !
Factoring the terms in brackets, collecting like terms, and isolating the
numbers to the rights hand side, we get in the explicit steps shown:
(x + 3)2 + (y + 1)2 + 6 = 9 + 1
(x + 3)2 + (y + 1)2 + 6 = 10
(x + 3)2 + (y + 1)2 = 4.
Thus: center = (-3, -1) and radius = 2. Note 22 = 4.
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