Algebra Assignment: Equations, Functions, and Applications

This document provides a comprehensive set of solutions to an algebra assignment, covering a wide range of topics. The solutions begin with basic concepts like evaluating functions and graphing, progressing to more complex topics such as exponential and logarithmic functions, including applications to population growth and carbon dating. The assignment explores various algebraic techniques for solving equations, interpreting graphs, and applying mathematical models to real-world scenarios. The solutions include detailed explanations, calculations, and graphs to illustrate the concepts. The document also includes references to relevant algebra textbooks. Overall, the assignment aims to provide students with a thorough understanding of algebraic principles and their practical applications.

Algebra
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Mathematics

Algebra
Contents
Solution-1).......................................................................................................................................3
Solution-2).......................................................................................................................................4
Solution-3).......................................................................................................................................5
Solution-3a).................................................................................................................................5
Solution-3b).................................................................................................................................6
Solution-3c).................................................................................................................................6
Solution-3d).................................................................................................................................7
Solution-4).......................................................................................................................................7
Solution-4a).................................................................................................................................7
Solution-4b).................................................................................................................................8
Solution-4c).................................................................................................................................8
Solution-4d).................................................................................................................................8
Solution-4e).................................................................................................................................9
Solution-5).......................................................................................................................................9
Solution-6).......................................................................................................................................9
Solution-7).....................................................................................................................................10
Solution-8).....................................................................................................................................11
Solution-8a)...............................................................................................................................11
Solution-8b)...............................................................................................................................11
Solution-9).....................................................................................................................................13
Solution-10)...................................................................................................................................13
Solution-11)...................................................................................................................................14
Solution-11a).............................................................................................................................14
Solution-11b).............................................................................................................................14
Solution-12)...................................................................................................................................14
Solution-12a).............................................................................................................................14
Solution-12b).............................................................................................................................15
Solution-13)...................................................................................................................................15
Solution-14)...................................................................................................................................16
Solution-15)...................................................................................................................................16
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Algebra
Solution-15a).............................................................................................................................16
Solution-15b).............................................................................................................................16
Solution-16)...................................................................................................................................17
Solution-17)...................................................................................................................................18
Solution-17a).............................................................................................................................18
Solution-17b).............................................................................................................................18
Solution-17c).............................................................................................................................18
Solution-17d).............................................................................................................................18
Solution-17e).............................................................................................................................18
Solution-17f)..............................................................................................................................19
Solution-18)...................................................................................................................................19
Solution-19)...................................................................................................................................20
Solution-20)...................................................................................................................................21
References......................................................................................................................................21
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Algebra
Solution-1)
As given in question,
h ( x )= (1
2 )x
The coordinate in x axis for given problem is given in graph, for x axis it is (-5, -4, -3. -1, 0, 1, 2,
3, 4, 5), we have to put the value of x in equation calculated the h(x), as result in the table is
given below.
x h(x)
-5 32
-4 16
-3 8
-2 4
-1 2
0 1
1 0.5
2 0.25
3 0.125
4 0.0625
5 0.03125
The graph for the above data will as follows
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-6 -4 -2 0 2 4 6
0
5
10
15
20
25
30
35
x vs h(x)

Algebra
Solution-2)
The calculated table for h(x) and g(x) is given below, the graph,
x h(x) g(x)
-5 32 65
-4 16 33
-3 8 17
-2 4 9
-1 2 5
0 1 3
1 0.5 2
2 0.25 1.5
3 0.125 1.25
4 0.0625 1.125
5 0.03125 1.0625
The graph for the above data will as follows
-6 -4 -2 0 2 4 6
0
10
20
30
40
50
60
70
x vs h(x) and g(x)
h(x)
g(x)
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Algebra
Solution-3)
As given in question,
The accumulated value of investment can be given by the formula
A=P (1+ r
n )nt
……….(i)
Solution-3a)
Where P = Principal,
r = rate per annum,
n = is compounding period for semi-annually = 2, Putting the other value in equation, we get
A=10000 ( 1+ 0.055
2 )
5 x 2
A=10000 (1.311651 ) =13116.51
The total value for semi-annually compounded is = $ 13116.51
Solution-3b)
For compounded quarterly the value of n will be =4, Putting then
P = 10000, r = 0.055, n = 4, and t = 5
Putting the value in above equation
A=10000 ( 1+ 0.055
4 )
5 x 4
A=10000 (1.314067 )=13140.67
For quarterly compounded is approximately $ 13140.67
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Algebra
Solution-3c)
Compounded monthly means we have to take n = 12
Putting the values in equation with n = 12
A=10000 (1+ 0.055
12 )5∗12
A=10000 (1.315704 )=13157.04
Therefore, total amount for monthly compounded interest is $ 13157.04
Solution-3d)
The equation for compounded continuously is given as
A=P ert ………. (ii)
We know that, P = 10000, r = 0.055, t = 5
Putting the value in above equation
A = 10000.e (0.055)5
A = 10000 x e0.275 = 13165.31
The accumulated value for the investment = 13165.31
Solution-4)
As given in question,
f ( x)=574(1.026)x This illustrates the population …….(i)
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Algebra
Solution-4a)
When x = 0, Putting the value of x in above equation
f (0)=574 (1.026)0
f ( 0 ) =574∗1= 574 million
Therefore, India’s population in 1974 is around 574 million
Solution-4b)
When x = 27, the nearest population will be as 2001
Putting the value x = 27
f (27)=574 (1.026)27
f ( 27 ) =574 x 1.9998=1147.87 million
Therefore, population in 2001 is around 1147.87 million
Solution-4c)
Year 2028 means after = 2028-1974 = 54 years from 1974
Putting the value of x = 54
f (54)=574 (1.026)54
f ( 54 )=574 x 3.999=1147.87 million
f ( 54 )=574 x 3.999=2295.46 million
Therefore, in 2028 the population will be 2295 million
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Algebra
Solution-4d)
Year 2055 means after = 2028-1974 = 81 years from 1974
Putting the value of x = 81
f (81)=574 (1.026)81
f ( 81 )=574 x 7.999=4590.374 million
f ( 81 )=574 x 7.999=4590.374 million
Therefore, in year 2055 the population will be 4590 million
Solution-4e)
As we can see, the population after 27 years i.e. in 2001 is 2294 million, which is around 4 time
of population in 1974, similarly again after 27 years the population is 4590 million, which is
around 8 time of the population in 1974, It can be concluded that the population of India is
growing 4 times after each 27 years.
Solution-5)
As given in question, 5=logb 32
Log is being defined as
For x > 0, and b > 0, and b ≠ 1
y=logb x is equivalent to by = x
Therefore putting the value place for y = 5 and x =32
We get
b5=32
Ans.
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Algebra
Solution-6)
As per the function g(x) = log(x-2), first we have to calculate different data of log(x-2) for given
value of x. The result is tabulated which is given below
x y
3 0.00
4 0.30
5 0.48
6 0.60
7 0.70
8 0.78
9 0.85
10 0.90
11 0.95
12 1.00
Now plotting the given function he curve obtained is as follows
3 4 5 6 7 8 9 10 11 12 13
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.30
0.48
0.60
0.70
0.78
0.85 0.90 0.95 1.00
x vs log(x-2)
Logarithmic function of the form
F(x) = c.log a ( x +h ) + k has vertical asymptotes x = -h
Therefore h = -2
The vertical asymptotes is x =2
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Algebra
F(x) = c.loga ( x +h ) +k Has no horizontal asymptote
The domain for log(x-2), : [ Solution : x> 2
Interval No tation :(2 , ∞) ] Ans
The Range for log(x-2): [ Solution :−∞<f ( x ) <∞
Interval Notation :(−∞ , ∞) ]
Solution-7)
As given in question,
The function of age percentage for girls is given as
f ( x )=62+ 35 log ( x−4)
Putting the value x = 13 we get
f ( 13 ) =62+35 log (13−4)
f ( 13 )=62+35 log (9)
f ( 13 )=62+35∗0.954243
f ( 13 )=62+33.39849
f ( 13 )=95.39849
F(13) = 95.40 % Ans
Solution-8)
As given in question
f ( t )=88−15 ln ( t +1 ) , 0<t <12
Solution-8a)
The original exam means t =0
f ( 0 )=88−15 ln ( 0+1 )
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Algebra
f ( 0 )=88−15∗0
f ( 0 )=88
Solution-8b)
After 2 month
f ( 2 ) =88−15 ln ( 2+1 )
f ( 2 ) =88−15 ln ( 2+1 )
f ( 2 ) =88−15 ln ( 3 )
f ( 2 ) =71.52082
After 4 month
f ( 4 )=88−15 ln ( 4+1 )
f ( 4 )=88−15 ln (5 )
f ( 4 )=88−24.1512
f ( 4 ) =63.85843
After 6 month
f ( 6 )=88−15 ln ( 6+1 )
f ( 6 )=88−15 ln ( 7 )
f ( 6 )=88−29.189
f ( 6 )=58.8113
After 8 month
f ( 8 )=88−15 ln ( 8+1 )
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Algebra
f ( 8 )=88−15 ln ( 9 )
f ( 8 ) =88−32.958
f ( 8 )=55.04163
After 10 month
f ( 10 ) =88−15 ln ( 10+ 1 )
f ( 10 )=88−15 ln (11 )
f ( 10 )=88−35.968
f ( 10 )=52.03157
After 12 month
f ( 12 ) =88−15 ln ( 12+1 )
f ( 12 ) =88−15 ln ( 13 )
f ( 12 ) =88−38.474
f ( 12 ) =49.52576
Solution-9)
log9 ( 9
x )can be expanded as follows
log9 ( 9
x ) ⟹ log9 9−log9 x
⟹ 1−log9 x Ans
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Algebra
Solution-10)
As we know that, the log rule is
log(a)−log(b)=log(a/b),
Substituting the respective value in given equation
The condensed for of log(2x+5) – logx
Is = log 2 x +5
x Ans
Solution-11)
As we know that,
As per log rule is log(a)−log (b)=log (a/b)
Solution-11a)
D=10 ¿
D=10 log I
log I0
⟹ D=10 log ( l
l0 ) Ans
Solution-11b)
As given in question,
The sound has intensity 100 times larger than softer sound
Therefore,
D = 10 log10 100
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Algebra
We know that x=logb y ⇒bx= y
Therefore, log10 100=x can be simplified as 10x = 100 or x = 2
Therefore, D = 10 x 2 = 20 decibel Ans.
Solution-12)
We know that,
As per log rule is log(a)−log (b)=log (a/b)
Solution-12a)
Using the above formula in the given equation
t= 1
c [ ln A−ln ( A−N ) ]
t= 1
c [ ln ( A
A−N ) ] This is the required single logarithm
Solution-12b)
As given in question,
C = 0.03, A = 65, N = 30
Putting the values in formula
t= 1
0.03 [ ln ( 65
65−30 ) ]
t= 1
0.03 [ 0.619039 ] = 20.63464 Ans
It is about 20.63 week will be taken by chimpanzee to master 30 sign
Solution-13)
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Algebra
As given in the question,
Exponential function is given as 9ex = 107
To separate the exponential function we have to divide both side equations by 9
9 ex
9 =107
9
ex= 107
9
Taking natural log on both side
ln e x=ln 107
9 (Since we know that ln e x=x
X = ln 107
9 = ln 11.89 = 2.47569771 = 2.48
The solution for given equation is ln 107
9 which is equal to 2.48
Solution-14)
As given in equation
ln √ x+ 3=1 (We know that a =logb ba)
1 = ln (e1) = ln (e)
l n ( √ x +3 ) =ln (e)
Removing ln from both side
√ x+ 3=e
x +3=e2 or, x = e2 – 3
x = 2.718282 – 3 = 7.389046 – 3 = 4.389046 Ans.
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Algebra
Solution-15)
As given in question,
A=37.3 e0.0095 t ……..(i)
Solution-15a)
To calculate the population for 2010 we have to take t = 0. Putting the value of t in equation (i)
A=37.3 e0.0095 t
A=37.3 e0.0095 x 0=37.3 e0
A=37.3 x 1= 37.3 million
Solution-15b)
In this problem we have to calculate time for being 40 million populations
40=37.3 e0.0095t
40 /37.3=e0.0095t
Taking ln from both side
Ln (40/37.3) = lne0.0095t
Ln 1.072386 = 0.0095t
T = 0.069886 / 0.0095 = 7.35 years
7.35 year from 2010 will April 2017 when population will reach to 40 million
Solution-16)
As given in question
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Algebra
f ( x )=20(0.975)x
To know the depth for 1% sunlight, we have to put f(x) = 1
1=20 (0.975)x
1
20 =(0.975)x
Taking log from both sides
log 0.05=xlog0.975
x= log 0.05
log 0.975
Or x = 118 ft.
By the calculated answer we can say that at the depth of 118 ft, the % of sunlight will be 1%
In the graph we can see that (118,1), 118 on x axis and 1 on y axis.
Solution-17)
As given,
Solution-17a)
The equation for Japan, A=127.3e–0.006t
Putting the value t = 0 for Japan
A = 127.3 x e-0.006x0 = 127.3 x 1 = 127.3 million
Solution-17b)
The equation for Iraq, A=31.5e0.019t
Again putting the value of t = 0
A=31.5e0.019x0 = 31.5 x e0 = 31.5
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Algebra
Solution-17c)
From the given all four equation we can see that Iraq and India has positive power, rest
has negative power. The power of Iraq is greater than India, which 0.019, Therefore growth rate
for Iraq is highest
Solution-17d)
From the equation, it clear that Russia and Japan has negative power to their equation, therefore
japan and Russia have negative growth rate
The power of Russia is -0.005, therefore its % of decrease will be 0.5 % / year whereas Japan has
power factor 0.006, so it is 0.6%.
Solution-17e)
The equation for India is
A=1173.1e0.008t Putting A = 1377
1377 / 1173.1 = e0.008t
Taking natural log from both side
Ln1.173813 = 0.008t
T = 0.16026/0.008 = 20.03
After approx. 20 years i.e. in 2030 the population of India will be 1377 million
Solution-17f)
The equation for India is
A=1173.1e0.008t Putting A = 1491
1491 / 1173.1 = e0.008t
Taking natural log from both side
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Algebra
Ln1.270991 = 0.008t
T = 0.239797/0.008 = 29.97465
After approx. 30 years i.e. in 2040 the population of India will be 1491 million
Solution-18)
The equation for carbon decay is
A=16 e−0.000121 t
Putting the value t = 5715
A=16 e−0.000121 t
A=16 e−0.000121 x 5715
A=16 e−0.691515
A = 8.013068 gm
Therefore after 5715 years the carbon -14 remains will be 8.013 years
Solution-19)
The decay model is given by the following equation, first we have to find the value of k through
half-life data.
A=A0 ekt
A0
2 = A0 ek(7340)
1
2 =ek(7340)
Taking natural log from both side
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Algebra
Ln0.5 = k*7340
-0.69315 = k x 7340
K = -0.69315 / 7340 = 0.0009443
The decay rate for Thorium-229 is k = -0.0009443 / year
Now again putting the value of k and calculating for 20 % remains of Thorium
0.2=e0.0009443 t
Taking Natural log from both side
ln 0.2=0.0009443t
-1.61 = 0.0009443 t
t= −1.61
0.0009443 =1704.37 years
Therefore, after 1704.37 year only 20% thorium will remain
Solution-20)
The given table is tabulated as given below
Age at death
Savings
needed
Differen
ce Ratio
80 219000
85 307000 88000
0.7133
55
90 409000 102000
0.7506
11
95 524000 115000
0.7805
34
100 656000 132000
0.7987
8
From the above table we can see the in difference column there is no similarity is visible. But in
ration column we can see that it is around 0.75, which is almost constant. The graph also show
some slight curve
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Algebra
75 80 85 90 95 100 105
0
100000
200000
300000
400000
500000
600000
700000
Savings needed Vs Age of death
So the equation will be in y =abx form which exponential. The equation will be y = 2896e0.0546x
Ans.
Bibliography
Blitzer, R. (2018). College Algebra (7th ed.). Newyork: Pearson Education,.
Kaufmann, J. E. (2014). Intermediate algebra (1st ed.). New York: Cengage Learning.
Larson, R. (2010). Intermediate Algebra (5th ed.). Newyork: Cengage Learning.
Mark Clark, C. A. (2012). Intermediate Algebra: Connecting Concepts through Applications (1st
ed.). Belmont: Cengage Learning.
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