𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔
1. Exponential/Logarithmic Equations. Solve for x in the following equations. Write your
answers as decimals numbers rounded to 3 digits after the decimal point.
(a) 𝑒 2𝑥−6 = 52
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑒 2𝑥−6 = 52;
𝑇𝑎𝑘𝑖𝑛𝑔 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑙𝑜𝑔𝑎𝑟𝑖𝑡ℎ𝑚𝑠𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠;
𝑙𝑛𝑒 2𝑥−6 = 𝑙𝑛52
(2x − 6)𝑙𝑛𝑒 = 𝑙𝑛52
(→ 𝑙𝑛𝑎𝑏 = 𝑏𝑙𝑛 𝑎)
𝑙𝑛52
(2𝑥 − 6) =
= 3.951243719
𝑙𝑛𝑒
𝑥=
3.951243719 + 6
= 4.975621859
2
𝒙 = 𝟒. 𝟗𝟕𝟔(𝟑 𝒅. 𝒑)
(b) 31𝑥 = 439
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
31x = 439
Taking logarithms on both sides;
log(31x ) = log 439
log ab = blog a
xlog 31 = log 439
x=
log439
= 1.771846851
log31
𝐱 = 𝟏. 𝟕𝟕𝟐 (𝟑 𝐝. 𝐩)
1
(c) log 3 (3𝑥 + 4) − log 3 (𝑥 + 1) = 2
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑙𝑜𝑔3 (3𝑥 + 4) − 𝑙𝑜𝑔3 (𝑥 + 1) = 2
2 = 𝑙𝑜𝑔3 9
log 3 (3𝑥 + 4) − log 3 (𝑥 + 1)
= 𝑙𝑜𝑔3 9
𝑏
𝑙𝑜𝑔𝑎 𝑏 − 𝑙𝑜𝑔𝑎 𝑐 = 𝑙𝑜𝑔𝑎 ( )
𝑐
3𝑥 + 4
𝑙𝑜𝑔3 (
) = 𝑙𝑜𝑔3 9
𝑥+1
3𝑥 + 4
=9
𝑥+1
3𝑥 + 4 = 9(𝑥 + 1)
3𝑥 + 4 = 9𝑥 + 9
5
−6𝑥 = 5 𝑥 = −
6
𝑥 = −0.833 (3 𝑑. 𝑝)
2. Growth and Decay. Imagine that you put $1500 into an account that pays 5.6% per
year, compounded continuously.
(a) What function gives the amount in the account after t years?
𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏;
𝒘𝒉𝒆𝒓𝒆 ;
𝒓 𝒕
𝑷 − 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒍𝒆 𝒂𝒎𝒐𝒖𝒏𝒕,
𝑨 = 𝑷 (𝟏 +
)
𝒓 − 𝒓𝒂𝒕𝒆
𝟏𝟎𝟎
𝒕 − 𝒕𝒉𝒆 𝒕𝒊𝒎𝒆 𝒊𝒏 𝒚𝒆𝒂𝒓𝒔
𝑨 − 𝑨𝒎𝒐𝒖𝒏𝒕 𝒂𝒇𝒕𝒆𝒓 𝒕𝒊𝒎𝒆 𝒕
(b) What is the balance after 5 years?
𝐴 = 𝑃 (1 +
𝑟 𝑡
)
100
5.6 5
𝐴 = 1500 (1 +
) = 1969.75
100
𝐻𝑒𝑛𝑐𝑒,
𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑎𝑓𝑡𝑒𝑟 5 𝑦𝑒𝑎𝑟𝑠 = $1969.75
𝑃 = 1500; 𝑟 = 5.6 ; 𝑡 = 5
(c) What is the balance after 10 years?
𝑟 𝑡
𝐴 = 𝑃 (1 +
)
100
5.6 10
𝐴 = 1500 (1 +
) = 2586.60
100
𝐻𝑒𝑛𝑐𝑒,
𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑎𝑓𝑡𝑒𝑟 10 𝑦𝑒𝑎𝑟𝑠 = $2586.60
𝑃 = 1500; 𝑟 = 5.6 ; 𝑡 = 10
2
(d) How long does it take for the account to double in value?
𝑟 𝑡
5.6 𝑡
𝐴 = 𝑃 (1 +
)
3000 = 1500 (1 +
)
100
100
3000
𝑡
(1
+
0.056)
=
=2
𝑃 = 1500; 𝑟 = 5.6 ;
1500
𝐴𝑚𝑜𝑢𝑛𝑡 = 2𝑥1500 = 3000
𝑡𝑎𝑘𝑖𝑛𝑔 𝑙𝑜𝑔𝑎𝑟𝑖𝑡ℎ𝑚𝑠
𝑙𝑜𝑔(1.056)𝑡 = log 2
𝑡𝑙𝑜𝑔1.056 = log 2
log 2
𝑡=
= 12.72
𝑙𝑜𝑔1.056
𝐻𝑒𝑛𝑐𝑒,
𝑡ℎ𝑒 𝑎𝑐𝑐𝑜𝑢𝑛𝑡 𝑤𝑖𝑙𝑙 𝑑𝑜𝑢𝑏𝑙𝑒 𝑎𝑓𝑡𝑒𝑟 12.72 𝑦𝑟𝑠
3. Growth and Decay. A rule of thumb used by car dealers is that the trade-in value of a
car decreases by a fixed percentage each year. A mini-cooper purchased in 2005 cost
$25,400. According to Kelly Blue book, this car could be sold privately $7,200 in 2013.
(a) Let t be the number of years since 2005. What is k, the decay rate, for this car?
𝑘 𝑡
𝐴 = 𝑃 (1 −
)
100
𝑤ℎ𝑒𝑟𝑒 ;
𝑃 − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑟
𝑘 − 𝑑𝑒𝑐𝑎𝑦 𝑟𝑎𝑡𝑒
𝑡 − 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑖𝑛 𝑦𝑒𝑎𝑟𝑠
𝐴 − 𝑉𝑎𝑙𝑢𝑒 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑡
𝐴 = 7200; 𝑃 = 25400;
𝑡 = 8 ( 2005 𝑡𝑜 2013)
𝑘 8
7200 = 25400 (1 −
)
100
7200
(1 − 0.01𝑘)8 =
= 0.283464566
25400
𝒕𝒂𝒌𝒊𝒏𝒈 𝒍𝒐𝒈𝒂𝒓𝒊𝒕𝒉𝒎𝒔 𝒐𝒏 𝒃𝒐𝒕𝒉 𝒔𝒊𝒅𝒆𝒔
𝑙𝑜𝑔(1 − 0.01𝑘)8 = 𝑙𝑜𝑔(0.283464566)
8𝑙𝑜𝑔(1 − 0.01𝑘) = 𝑙𝑜𝑔(0.283464566)
𝑙𝑜𝑔(0.283464566)
𝑙𝑜𝑔(1 − 0.01𝑘) =
8
𝑙𝑜𝑔(1 − 0.01𝑘) = −0.068437652
𝒕𝒂𝒌𝒊𝒏𝒈 𝒂𝒏𝒕𝒊𝒍𝒐𝒈𝒂𝒓𝒊𝒕𝒉𝒎𝒔 𝒐𝒏 𝒃𝒐𝒕𝒉 𝒔𝒊𝒅𝒆𝒔
1 − 0.01𝑘 = 0.8542;
1 − 0.8542
𝑘=
= 14.579
0.01
𝑯𝒆𝒏𝒄𝒆 𝒕𝒉𝒆 𝒅𝒆𝒄𝒂𝒚 𝒓𝒂𝒕𝒆 = 𝟏𝟒. 𝟓𝟕𝟗%
3
(b) What was the value of the car in 2010?
𝑘 𝑡
𝐴 = 𝑃 (1 −
)
100
𝑃 = 25400; 𝑟 = 14.579; 𝑡 = 5
14.579 5
𝐴 = 25400 (1 −
)
100
𝐴 = 25400𝑥0.854215 =11551.99
𝑯𝒆𝒏𝒄𝒆, 𝒗𝒂𝒍𝒖𝒆 𝒊𝒏 𝟐𝟎𝟏𝟎 𝒘𝒂𝒔 $𝟏𝟏𝟓𝟓𝟏. 𝟗𝟗
1
(c) How long does it take for the car to be worth 10 of its original value?
𝐴 = 𝑃 (1 −
𝑘 𝑡
)
100
14.579 𝑡
)
100
2540
(1 − 0.14579)𝑡 =
= 0.1
25400
𝑡𝑎𝑘𝑖𝑛𝑔 𝑙𝑜𝑔𝑎𝑟𝑖𝑡ℎ𝑚𝑠
2540 = 25400 (1 −
𝑃 = 25400;
𝑘 = 14.579
1
𝐴=
𝑥 25400 = 2450
10
𝑙𝑜𝑔(0.85421)𝑡 = log 0.1
𝑡𝑙𝑜𝑔(0.85421) = log 0.1
log 0.1
𝑡=
= 14.61
log 0.85421
𝐻𝑒𝑛𝑐𝑒,
𝑖𝑡 𝑤𝑖𝑙𝑙 𝑡𝑎𝑘𝑒 14.61𝑦𝑟𝑠
4
9
4. Acute Angles. Given sec 𝛼 = 7 , find each of the following. Note: write your answers
as decimals rounded to 2 digits after the decimal point. Be sure to include units as
degrees or radians where appropriate! No need to simplify values in square root.
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
sec 𝛼 =
1
9
=
cos 𝑎
7
7
9
2
√
𝑎 = 9 − 72 = √14
𝑏 = 7; 𝑐 = 9
cos 𝛼 =
7
𝛼 = cos −1 ( ) = 38.94°
9
a = √14
sin 𝛼 =
7
𝛽 = sin−1 ( ) = 51.06°
9
𝑏=7
√14
9
= 0.42
1
cot 𝛼 = tan 𝛼 =
7
√14
cos 𝛼 =
𝑐 =9
7
9
= 0.78
1
tan 𝛼 =
9
√14
7
= 0.53
1
= 1.87 sec 𝛼 = 𝑐𝑜𝑠𝛼 = 7 = 1.29 csc 𝛼 = sin 𝑎 =
9
√14
= 2.41
5. Right Triangles. In one area, the lowest angle of elevation of the sun in winter is
27.35°. Find the minimum distance x that a plant needing full sun can be placed from a
fence that is 5.7 feet high. Round your answer to the tenths place
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑜𝑝𝑝
;
𝑎𝑑𝑗
∅ = 27.35;
𝑜𝑝𝑝 = 5.7;
tan ∅ =
𝑥 = 𝑎𝑑𝑗 =
𝑜𝑝𝑝
5.7
=
= 11.01994145
tan ∅ tan 27.35
𝑥 = 𝟏𝟏. 𝟎 𝒇𝒆𝒆𝒕
5
6. Given cot 𝜃 = −1 and 540° ≤ 𝜃 ≤ 720°
(a) What is 𝜃?
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛;
cot 𝜃 = −1 =
1
= −1; 𝑡𝑎𝑛𝜃 = −1
𝑡𝑎𝑛𝜃
𝜃 = 315 + 360𝑛
n
0
1
315
675
𝜃
𝜃
= −45 (𝜃 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 2𝑛𝑑 𝑎𝑛𝑑 4𝑡ℎ 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 )
𝐼𝑛 𝑡ℎ𝑒 4𝑡ℎ 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡;
𝜃 = 360 − 45 = 315°.
𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙;
𝜃 = 315 + 360𝑛;
𝑤ℎ𝑒𝑟𝑒 𝑛 = 0,1,2,3 …
(b) Sketch the angle on a coordinate system.
6
𝐹𝑜𝑟 540° ≤ 𝜃 ≤ 720°
𝜽 = 𝟔𝟕𝟓°
2
1035
7. Large Angles. Use the picture below to fill in the values below. Note: write your
answers as decimals rounded to 3 digits after the decimal point.
tan 𝑥 =
3
;
4
𝑥 = tan−1(0.75) = 36.870°
𝜃 = 𝑥 + 360 = 396.870
𝜃 = 396.870°
sin 𝜃 = −0.600
cot 𝜃 = 1.333
cos 𝜃 = − 0.800
sec 𝜃 = −1.250
tan 𝜃 = 0.750
csc 𝜃 = −1.667
8. Complete the table below.
s
r
𝜽
0.92
2.4
0.38 radians
25.32
48.35
𝜋
6
0.96
3.1
0.31 radians
7
8