Find x. Remember that if we know the base, there are only two other numbers. One number is the logarithm, and the other number is the antilogarithm. Thus we will always use the LOG or LN key or the INV LOG or INV LN keys.
a. x = ln 0.0052
x = -5.26.
b. e^x = 51.4
x = 3.94.
c. ln x = -4.16
x = e^-4.16
x = 0.016.
d. log x = -4.16
x = 10^-4.16
x = 0.000069.
e. Use the LN key to express each number as a power. Then use the rules of exponents to find the answer.
(0.000612)(576) / (0.0512 * 10^-14)
(e^-7.40)(e^6.36) / (e^-2.97)(e^-32.24)
e^(-7.40 + 6.36 + 2.97 + 32.24)
e^-34.17
6.89 * 10^14.
1. There were 19 nickels, dimes, and quarters in the pot. James noted that their value was $2. How many coins of each type were there if there were twice as many nickels as dimes?
a: 25NQ + 10ND + 5NN = 200
b: NQ + ND + NN = 19
c: NN = 2ND
a': 25NQ + 20ND = 200
b': NQ + 3ND = 19
25b': 25NQ + 75ND = 475
-a': -25NQ - 20ND = -200
55ND = 275
ND = 5
c: NN = 2(5)
NN = 10
b: NQ + 5 + 10 = 19
NQ = 4
10 nickels, 5 dimes, 4 quarters.
(Check: 25(4) + 10(5) + 5(10) = 100 + 50 + 50 = 200.)
2. Reds varied directly as blues and inversely as mauves squared. When there were 10 reds, there were 2 mauves and 4 blues. How many blues were there when there were 20 mauves and 3 reds?
R = kB/M^2
10 = k4/4
k = 10
3 = 10B/20^2
3 = 10B/400
10B = 1200
B = 120.
3. It was necessary to mix 200 mL of a solution in which the key ingredient made up exactly 63 percent of the total. One container held a solution that contained 70% key ingredient, and the other container held a solution that was only 60% key ingredient. How much of each solution should be used?
6A60 + 7A70 = 6.3(A60 + A70)
A60 + A70 = 200
6A60 + 7A70 = 1260
6A60 + 6A70 = 1200
A70 = 60
A60 = 140
60 mL of 70%; 140 mL of 60%.
4. Carlene made the trip in only 8 hours, whereas Dan took 12 hours to make the same trip. This was because Dan dawdled and drove 20 miles per hour slower than Carlene drove. How fast did each drive, and how long was the trip?
8RC = 12RD
RD = RC - 20
8RC = 12RC - 240
4RC = 240
RC = 60
RD = 40
D = 480
Carlene: 60 mph; Dan: 40 mph; trip: 480 miles.
5. The pressure of a quantity of an ideal gas was held constant at 1400 torr. The initial volume and temperature were 1000 mL and 1700 K. If the volume were increased to 2000 mL, what would the final temperature be in kelvins?
Doubled volume; double temperature: 3400 K.
Find x. Remember that in a logarithm problem we find either a logarithm or an antilogarithm.
log (base, "ln" = e "log" = 10) (antilogarithm) = logarithm. (to remember: log yields a logarithm)
base ^ logarithm = antilogarithm
6. (a) x = ln 0.0093: x = -4.68,
(b) e^x = 62.5: x = 4.14.
7. (a) ln x = 5163: x = e^5163,
(b) log x = 2.136: x = 136.77.
8. In the figure shown, AE = 6, AD = 4, and AB = 24. What is the length of the segment AC? Dimensions are in meters.
we have to rotate the smaller triangle but I HAVE NO IDEA HOW. :b
i mean ostensibly the solution is to just rotate it differently than i did before, and theres only one other way to do that, but......how should i know thats correct
(this is an even question so theres no answer in the back of the book)
pretty sure this is the first question like this, and not convinced the book ever suggested how :D
okay i think we can just use the hypotenuse as reference? maybe?
AE * SF = AB
no, that gets us where we were before - no closer to knowing whether AD corresponds to AC or CB
but corresponding to AC didnt work sooooo lets try CB
AD * SF = CB
only one other relationship, so:
ED * SF = AC
AE = 6
AD = 4
6^2 - 4^2 = 36 - 16 = 20 = ED^2
ED = sqrt20 = 2sqrt5
so whats the scale factor?
AE * SF = AB
6 * SF = 24
SF = 4
AC = ED * SF
AC = 2sqrt5 * 4
AC = 8sqrt5
we can check this by getting CB and then seeing if the numbers add up, pythagorean theorem-wise
CB = AD * SF
CB = 4 * 4
CB = 16
AC^2 + CB^2 = AB^2
(8sqrt5)^2 + 16^2 = 24^2
320 + 256 = 576
576 = 576
okay thats correct! so:
AC = 8sqrt5.
alternatively, assume: ADE is similar to ACB.
AD * SF = AC
AE * SF = AB
6 * SF = 24
SF = 4
4 * 4 = AC
AC = 16
CB = ED * SF
CB = 2sqrt5 * 4
CB = 8sqrt5
...this gets us to the same point
AC^2 + CB^2 = AB^2
16^2 + (8sqrt5)^2 = 24^2
256 + 320 = 576
576 = 576
okay, thats ALSO correct! ...so:
AC = 16
so AC = 16 or 8sqrt5, depending on how the triangles are similar.
this feels like an incomplete answer, but i dont see a way to tell how theyre similar?
AED is similar to ABC by A -> A, D -> C, so the only other relationship is E -> B
"AED is similar to ABC" is the same as "ADE is similar to ACB" so well take the "alternative" answer:
AC = 16.
Graph on a number line:
9. (x + 2)(x - 3) < 0; D = {Reals}.
x + 2 > 0 and x - 3 < 0
x > -2 and x < 3
(full solution)
or:
x + 2 < 0 and x - 3 > 0
x < -2 and x > 3
(no solution)
okay! its apparently normal for one resulting conjunction to have no solution, and to get the full solution from the other.
x > -2 and x < 3; D = {Reals}.
10. x^2 - x - 6 >= 0; D = {Integers}.
x^2 - x - 6 >= 0
(x + 2)(x - 3) >= 0
x + 2 > 0 and x - 3 > 0
x > -2 and x > 3
x > 3
or:
x + 2 < 0 and x - 3 < 0
x < -2 and x < 3
x < -2
x < -2 or x > 3; D = {Integers}.
11. x^2 - 8 <= 2x; D = {Integers}.
x^2 - 2x - 8 <= 0
(x + 2)(x - 4) <= 0
x + 2 > 0 and x - 4 < 0
x > -2 and x < 4
-2 < x < 4; D = {Integers}.
12. |x| - 1 !<= 0; D = {Integers}.
|x| !<= 1
|x| > 1
x < -1 or x > 1; D = {Integers}.
13. 7 !> x - 2 < 10; D = {Reals}.
7 <= x - 2 < 10
9 <= x < 12; D = {Reals}.
14. Multiply: (x^1/3 + y^2/3)(x^2/3 + y^1/3).
x^1/3 * x^2/3 + x^1/3 * y^1/3 + y^2/3 * x^2/3 + y^2/3 * y^1/3
x + x^1/3y^1/3 + x^2/3y^2/3 + y.
15. Factor: 8p^6k^15 - x^3m^6.
(2p^2k^5)^3 - (xm^2)^3
(a^3 - b^3) / (a - b)
a^3/a = a^2; *-b = -a^2b
a^2b/a = ab; *-b = -ab^2
ab^2/a = b^2; *-b = -b^3
(a - b)(a^2 + ab + b^2)
(2p^2k^5 - xm^2)(4p^4k^10 + 2p^2k^5xm^2 + x^2m^4).
16. Show that 0.003|16 is a rational number by writing it as a fraction of integers.
100N = 0.316|16
N = 0.003|16
99N = 0.313
99000N = 313
N = 313/99000.
17. Complete the square as an aid in graphing:
y = -x^2 + 4x - 1 opens downward; y-intercept = -1
y = -(x - 2)^2 + 3 axis = 2; vertex = 3
18. Find the number that is 2/11 of the way from 2 to 4 1/6.
2 + 2/11 * (4 1/6 - 2)
2 + 2/11 * (2 1/6)
2 + 2/11 * 13/6
2 + 1/11 * 13/3
2 + 13/33
79/33.
19. Solve 3x^2 - x - 7 = 0 by completing the square.
x^2 - 1/3x = 7/3
-1/3 / 2 = -1/6; ^2 = 1/36
x^2 - 1/3x + 1/36 = 7/3 + 1/36
(x - 1/6)^2 = 85/36
x - 1/6 = +-sqrt(85)/6
x = 1/6 +- sqrt(85)/6.
Solve:
20.
a: (1+1/5)x + 2/3y = 30
b: -0.18x - 0.02y = -3.78
a': 18x + 10y = 450
b': 18x + 2y = 378
a': 18x + 10y = 450
-b': -18x - 2y = -378
8y = 72
y = 9
b': 18x + 2(9) = 378
18x = 360
x = 20
(20, 9).
21.
x^2 + y^2 = 4
3x - y = 2
y = 3x - 2
y^2 = 9x^2 - 12x + 4
10x^2 - 12x + 4 = 4
10x^2 - 12x = 0
5x^2 - 6x = 0
x^2 - 6/5x = 0
-6/5 / 2 = -3/5; ^2 = 9/25
x^2 - 6/5x + 9/25 = 9/25
(x - 3/5)^2 = 9/25
x - 3/5 = +-3/5
x = 3/5 +- 3/5
x = 0, 6/5
(0, -2), (6/5, 8/5).
22.
a: x - 4y = -15
b: 3x + z = 20
c: 2y - z = 5
b: 3x + z = 20
c: 2y - z = 5
d: 3x + 2y = 25
a: x - 4y = -15
2d: 6x + 4y = 50
7x = 35
x = 5
a: 5 - 4y = -15
-4y = -20
4y = 20
y = 5
c: 10 - z = 5
-z = -5
z = 5
(5, 5, 5).
23.
a: x - 2y - z = -8
b: 3x - y - 2z = -5
c: x + y + z = 9
a: x - 2y - z = -8
c: x + y + z = 9
d: 2x - y = 1
b: 3x - y - 2z = -5
2c: 2x + 2y + 2z = 18
e: 5x + y = 13
d: 2x - y = 1
e: 5x + y = 13
7x = 14
x = 2
d: 4 - y = 1
-y = -3
y = 3
c: 2 + 3 + z = 9
z = 4
(2, 3, 4).
Simplify:
24. (2i^3 - 1) / (-sqrt(-3)sqrt(-3) + 3)
(-1 - 2i) / 6
-1/6 - 2/6i
-1/6 - 1/3i.
25. (sqrt2 - 5) / (2sqrt2 - 4)
(sqrt2-5)(2sqrt2+4) = -16 - 6sqrt2
(2sqrt2-4)(2sqrt2+4) = -8
(-16 - 6sqrt2) / -8
(16 + 6sqrt2) / 8
(8 + 3sqrt2) / 4.
26. (xy^3)^(1/6) * (xy^2)^(1/3)
x^(1/6 + 1/3) * y^(1/2 + 2/3)
x^1/2y^7/6.
27. sqrt(4/3) + 2sqrt(3/4) - 3sqrt48
sqrt4/sqrt3 + 2sqrt3/sqrt4 - 3sqrt48
2/sqrt3 + sqrt3 - 3sqrt48
2sqrt3/3 + sqrt3 - 12sqrt3
(2/3 + 1 - 12)sqrt3
(2 + 3 - 36)sqrt3/3
-31sqrt3/3.
Solve by factoring:
28. 2x^3 = -x^2 + 3x
2x^3 + x^2 - 3x = 0
x(2x^2 + x - 3) = 0
(2)(-3) = -6; 2 - 3 = -1
(3)(-2) = -6; 3 - 2 = 1
x(2x^2 - 2x + 3x - 3) = 0
x(2x(x - 1) + 3(x - 1)) = 0
x(2x + 3)(x - 1) = 0
x = 0; x = 0
2x + 3 = 0; 2x = -3; x = -3/2
x - 1 = 0; x = 1
x = 0, -3/2, 1.
29. 3x^2 - 2 = x
3x^2 - x - 2 = 0
(3)(-2) = -6; 3 - 2 = 1
(2)(-3) = -6; 2 - 3 = -1
3x^2 - 3x + 2x - 2 = 0
3x(x - 1) + 2(x - 1) = 0
(3x + 2)(x - 1) = 0
3x + 2 = 0; 3x = -2; x = -2/3
x - 1 = 0; x = 1
x = -2/3, 1.
30. -7x^2 - 2x = 3x^3
3x^3 + 7x^2 + 2x = 0
x(3x^2 + 7x + 2) = 0
(3)(2) = 6; 3 + 2 = 5
(1)(6) = 6; 1 + 6 = 7
x(3x^2 + 6x + 1x + 2) = 0
x(3x(x + 2) + 1(x + 2)) = 0
x(3x + 1)(x + 2) = 0
x = 0; x = 0
3x + 1 = 0; 3x = -1; x = -1/3
x + 2 = 0; x = -2
x = 0, -1/3, -2.