Algebra Logarithms Solve

The algebra logarithms solve represents the problems in logarithms that uses the algebra. The logarithm for the given number to a given base is generally the power or exponent to which the base must be raised in order to produce that number.Sometimes the logarithmic table  are used in the problems for evaluating the values in the forms of logarithmic algebra.


Examples to explain algebra logarithms solve

solve the natural logarithm for ln ((ab)2/ca)

solution:

ln ((ab)2/ca)

=ln ((ab)2) -  ln (ca)    by property Quotient formula

=ln a2 + ln b2  - (ln c+ ln a)  by property Product formula

=2 ln a + 2 ln  b- ln c- lna

= ln a +  2 ln b - ln c

Consider the same problem without using natural logarithms.

solve log ((ab)2/ca)

=log ((ab)2) -  log (ca)    by property Quotient formula

=log a2 + log b2  - (log c+ log a)  by property Product formula

=2 log a + 2 log  b- log c- log a

= log a +  2 log b - log c

solve the following log x expansions for logarithmic algebra  log1213 + 13 log2222



Given:

log12 13 + 13 log22 22

= log1213 + 13 (1)          ( log aa = 1)

= log1213 + 13

This is the required log x expansion. Hence we solved the problem.

Problems to explain algebra logarithms solve

Solve log51003

Solution:

log51003=3 log5100

= 3 log5(10*10)

= 3 (log510 + log510)

= 3(2 log510)

Log51003= 6 log510.

Solve the following log x expansions for logarithmic algebra log17y -  log17 x +  log1715

Solution:

log17 y - log17 x + log17 15              (given)

= log17`(y / x)` + log17 15                     ( log a - log b = log `a/b` )

= log17( `y/x`  . 15)                                (log a + log b = log a.b)


This is the required log x expansion. Hence we solved the above problem.

Prove that  log3 4 x log4 5 x log5 6 x log6 7 x log7 8 x log8 9 = 2

Solution:
Left Hand Side = ( log3 4 x log4 5 ) x ( log5 6 x log6 7 )  x ( log7 8 x log8 9  )
=log3 5  x  log5 7  x  log7 9
= log3 5 x  (  log5 7  x  log7 9)
= log3 5 x log5 9
= log3 9 = log3 32
= 2log3 3
= 2 x 1
= 2 = Right Hand Side
This problem explains the change of base rule.

Linear Algebra Vector

Linear algebra vector is one of the most important topics in linear algebra or Algebraic theory. In this linear algebra having some of the sub topics, like vector, matrix and so on.  This linear algebra having some more important application in matrices and vector algebra. Now we are going to see  about vector algebra and vector spaces in linear algebra theory.

Vector space in Algebra:

A vector space is a concept of a group of vectors or set of vectors. Which are working on two operations, like1.Vector addition and 2. Scalar multiplication. This is general concept of vector. And it is having more subtopics: Vector spaces, sub spaces, fundamental sub spaces, inner product spaces, span, basis and dimension, change of basis, linear independence, least squares, orthogonal matrices, QR- Decomposition, and orthonormal basis.

Example for vector space: the complex number 4 + 5i can be considered a vector space, it is way of explained by

`[[4],[5]]`

the vector space is a “space”, such as abstract objects.  Which is we called as vectors. Now we are study about the vectors, and see the vectors like S2, S3… Sn and so on. These all are like a vector spaces.
Vector addition is denoted by = x, y `in`  V, (x + y).


Properties of vector algebra:

In this vector addition and scalar multiplication is having some important properties, there are given below,

Vector addition property:

Commutative addition property:` vecx` + `vecy` = `vecy + vecx.`
Associative addition property: `vecx + (vecy + vecz)` = `(vecx + vecy) + vecz` .
Addition identity property: Here it is a 0 vector, so 0 +` vecx ` = `vecx `   for all x.
Addition inverse property: for each x vector, there exist another y vector like`vecx + vecy` = 0.



Scalar multiplication property:

Scalar associative property: α (βx) = (vecαβ) x.
Scalar distributive property: (α + β) x = αx+βx.
Vector distributive property: α(x+y) = αx+αy.
Scalar identity property: 1x = x.

These all are the very important in linear algebra properties in vector.


Types of vector:

Types of Vectors

Zero Vector A vector whose initial and terminal points coincide, is called a zero Vector (or null vector), and denoted as `vec0` . Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors `vec(A A) `, `vec(BB)` represent the zero vectors,

Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The Unit vector in the direction of `veca ` given vector a is denoted by `hata` .

Co initial Vectors Two or more vectors having the same initial point are called co initial vectors.

Collinear Vectors Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.

Equal Vectors Two vectors `veca` and `vecb` are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as `veca = vecb`

Negative of a Vector A vector whose magnitude is the same as that of a given vector (Say, `vec(AB)` ), but direction is opposite to that of it, is called negative of the given vector.

For example, vector `vec(BA)` is negative of the vector `vec(AB)` , and written as `vec(BA) = vec(-AB)`

Trinomial Squaring Twice

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.(source : WIKIPEDIA)
The trinomial must be one of the following form .
Examples for trinomial:    
1.  x + y + z , where x , y, z are variables.
2. 2xy + 3x + 4y, Where x, y are variables.
3. x²+2x-7, where x is variable.
4. ax + by + c = 0 , Where a,b,c are constants and x,y are variables.
In the following section we are going to see some solved problems and practice problems on trinomial squaring twice .

Solved problems on trinomial squaring twice:

Trinomial squaring twice problem 1:
Solve by squaring twice (2x + 5y + 6)⁴
Solution:
Given , (2x + 5y + 6)⁴
We can write it as , (2x + 5y + 6)⁴  = (2x + 5y + 6)^{2^2}               
                                                             = (2x + 5y + 6)² × (2x + 5y + 6)²
Let us take (2x + 5y + 6)²
We can apply the following formula to square the trinomial,
(a+b+c)² = ( a² + b² + c² + 2ab + 2bc + 2ca )
(2x + 5y + 6)² = (4x² + 25 y² + 36 + 20xy + 60x + 24x)
(2x + 5y + 6)² × (2x + 5y + 6)² = (4x² + 25 y² + 36 + 20xy + 60x + 24x) × (4x² + 25 y² + 36 + 20xy + 60x + 24x)
Answer: 2x + 5y + 6)⁴ =   (4x² + 25 y² + 36 + 20xy + 60x + 24x) × (4x² + 25 y² + 36 + 20xy + 60x + 24x) 

Trinomial squaring twice problem 2:
Find the trinomial by squaring (3x +2).
Solution:
Given (3x +2)
We can get the trinomial by squaring the given binomial.
(3x+2)² = (3x + 2) ( 3x + 2)
              = 3x(3x + 2) + 2( 3x+2 )
              = 9x² + 6x + 6x + 4
              = 9x² + 12x + 4
Answer: (3x+2)² =  9x² + 12x + 4

Trinomial squaring twice problem 3 :
Factor the trinomial 2x²- 22 x + 48
Solution:
Given 2x²- 22 x + 48
Divide by 2 ,We get  x²- 11x + 24
                                                                     24   (product)
                                                                     /  \    
                                                                 - 8   - 3 
                                                                    \   / 
                                                                    - 11     (sum)
Factors:  (x-8) and (x-3).
Verification:
(x-8) (x-3)  = x(x-8) -3(x-8)
                = x²-8x -3x+24
(x-8) (x-3)  = x² -11x + 24

Practice problems on trinomial squaring twice:

Problems:
1. Find the trinomial by squaring (5x + 6 )
2.Find the trinomial by squaring (x-8)
Answers:
1.25x² + 60x + 36
2. x² -16x + 64

Study Correlation Coefficient

In this article we will discuss about study correlation coefficient. The correlation coefficient, an idea from statistics calculates of how well trends in the expected values follow trends in past real values. The correlation coefficient is a value between 0 and 1. If there is no association between the predicted values and the definite values the correlation coefficient is 0 or very low.

Formula for study correlation coefficient:

`"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

Where
              N = Total number of values
              X = 1st achieve
              Y = 2nd achieve
            sum XY = Addition of the product of 1st and 2nd achieve
              sum X = Addition of 1st achieve
              sum y = Addition of 2nd achieve
              sum x² = Addition of square 1st achieve
              sum y² = Addition of square 2nd get achieve

Example problems for study correlation coefficient:

Study Correlation coefficient – Example 1:
Calculate the Correlation co-efficient of following table.

X	60	61	62	63	64
Y	3.1	3.6	3.8	4	4.1

Solution 1 for study correlation coefficient:
            Step 1:  Count the number of values.
                         N = 5
            Step 2:  Calculate XY, X², Y²
                         See the below table

X	Y	x*y	x*x = x^2	y*y = y^2
60	3.1	60*3.1=186	60*60=3600	3.1*.31=9.61
61	3.6	61*3.6=219.6	61*61=3721	3.6*3.6=12.96
62	3.8	62*3.8=235.6	62*62=3844	3.8*3.8=14.44
63	4	63*4   =252	63*63=3969	4*4 = 16
64	4.1	64*4.1=262.4	64*64=4096	4.1*4.1=16.81

       Step 3:

       Find sumX, sumy, sumxy, sumx², sumy².

        Sum x = 310
        sum y = 18.6
        sum xy = 1155.6
        sum x² = 19230
        sum y² = 69.82

      Step 4:

       Now, Substitute in the above formula given.

            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

             = `[5(1155.6) - ((310)(18.6)) / ((sqrt([5(19230)-(310)^2][5(69.82)-(18.6)^2]))]]`
              = `5778 - 5766 / sqrt(([96150 - 96100] [349.1-345.96]))`
              = `12 / sqrt(50 x 3.14 )`
              = `12/ sqrt(157)`
            = `12/12.5299`
            = 0.9577
            Answer is 0.9577
Study Correlation coefficient – Example 2:
         Find the Correlation co-efficient of following table

X	Y
50	3
51	3
52	3
53	4
54	4

        


 Solution for study correlation coefficient:
        Step 1: Count the number of values.
            N = 5
        Step 2: Find XY, X², Y²
            See the below table

X value	Y value	x* y	x*x	y*y
70	3	70*3=210	70*70=4900	3*3=9
71	3	71*3=213	71*71=5041	3*3=9
72	3	72*3=216	72*72=5184	3*3=9
73	4	73*4=292	73*73=5329	4*4=16
74	4	74*4=296	74*74=5476	4*4=16

          Step 3: Find sum X, sum y, sum xy, sum x², sum y².
            sum x = 360  
            sum y = 17
            sum xy = 1227
            sum x² = 25930
            sum y² = 59
           Step 4: Now, Substitute in the above formula given.
            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`
              = `[5(1227) - ((360)(17)) / ((sqrt([5(25930)-(360)^2][5(59)-(17)^2]))]]`                
              = `(6135 - 6120)/((sqrt([129650 -129600] [295-289])))`
              = 1`5/sqrt(50 xx6)`
              = `15/sqrt(300) `
              = `15/17.32`
              = 0.8660
              Answer is: 0.8660

Least Common Multiple of 8

In mathematics, the least common multiple of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder. (Source: From Wikipedia).

Least common multiple of two numbers can be found by their multiples. Here we are going to learn how to find the least common multiple of two or more numbers.

Example problems to find the least common multiple of two numbers

Here we will see some example problems to find the least common multiple of two numbers.
Example 1
Find the least common multiple of 6 and 8
Solution
The least common multiple of 6 and 8 can be found by finding the multiples 6 and 8.
The list of multiples of 6 and 8 are given below
The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 72, 78, 84.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.
Here 24 is the lowest common number. So, 24 is the lowest common multiple of 6 and 8.
Example 2
Find the least common multiple of 5 and 8
Solution
The least common multiple of 5 and 8 can be found by finding the multiples 5 and 8.
The list of multiples of 5 and 8 are given below
The multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.

Here 40 is the lowest common number. So, 40 is the lowest common multiple of 5 and 8.
Example 3
Find the least common multiple of 7 and 8
Solution
The least common multiple of 7 and 8 can be found by finding the multiples 7 and 8.
The list of multiples of 7 and 8 are given below
The multiples of 7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.
Here 56 is the lowest common number. So, 56 is the lowest common multiple of 7 and 8.
Example 4
Find the least common multiple of 11 and 8
Solution
The least common multiple of 11 and 8 can be found by finding the multiples 11 and 8.
The list of multiples of 11 and 8 are given below
The multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88 , 99, 110
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88.
Here 88 is the lowest common number. So, 88 is the lowest common multiple of 11 and 8.

Practice problems to find the least common multiple of two numbers

Practice problem 1

Find the least common multiple of 10 and 8
Answer: The least common multiple of 10 and 8 is 40
Practice problem 2
Find the least common multiple of 12 and 8
Answer: The least common multiple of 12 and 8 is 24
Practice problem 3
Find the least common multiple of 15 and 8
Answer: The least common multiple of 15 and 8 is 120

Prime Number

• In mathematics, a prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself. The first twenty-five prime numbers are: Prime Numbers:- 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

• Condition: - If X is the prime number next the factors of the number x is 1 and X. (Source:Wikipedia)                                                                                                                                           

Prime number of 7?

• The prime number 7 has only two divisors 1 and 7.
• Here we are going to discuss about some prime number problems.

Example problems of prime number:

Example: - 1

          Find the number 17 is the prime number or not?

Solution:-

          Given number is 17 we have to find out it is prime number or not.

          The factors of the number 17 is the 1, 17 only.

          In the number 17 have only two factors so it is a prime number.

Example: - 2

          Find the number 331 is prime number or not?

Solution:-

         Given number is 331 we have to find out it is prime number or not.

          The factors of the number 331 is the 1, 331 only.

          In the number 331 have only two factors so it is a prime number.

Example:-3

        Find the number 1259 is prime number or not?

Solution:-

          Given number is 1259 we have to find out it is prime number or not.

          The factors of the number 1259 is the 1, 1259 only.

          In the number 1259 have only two factors so it is a prime number.

Example: - 4

        Find out the given number 8 is prime number or not?

Solution:-

        Given number is 8 we have to find out it is prime number or not.

        The factors of the number 8 are 1, 2, 4, and 8 so it is more than two.

        In the number 8 has two or more factor so it not a prime number.

Example: - 5

        Find the number 300 is prime number or not?

Solution:-

        Given number is 300 we have to find out it is prime number or not.

        The factors of the number 300 are 1, 3, 5, 15, 30 and 300 is more than two.

        In the number 300 have more than two factor hence it is not a prime number.

Ellipse Parabola Hyperbola

An ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve.

A parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point and a corresponding line on the plane, the locus of points in that plane that are equidistant from them is a parabola.

A hyperbola is a curve, specifically a smooth curve that lies in a plane, which can be defined either by its geometric properties or by the kinds of equations for which it is the solution set.(Source: Wikipedia)



Example ellipse problem :

Find the equation of ellipse with its center at the origin and having foci at (`+- 2sqrt(6)`, 0 )and an eccentricity

equal to the  ,`(2sqrt(6))/(7)`

Solution:

The focal points on the X axis is the ellipse is oriented, view A, and the standard form of the equation is ` (x^2)/(a) + (y^2)/(b)`

The center of the origin, the numerators of the fractions on the left is x2 and y2. The problem is find the values of a and b.

The distance from the center to be either of the foci is equal to c. so in this problem

C = `2sqrt(6)`

from the given coordinates of the foci.

The values of a, c, and e (eccentricity) are related by

c=a*e

or

` a = (c)/(e)`

Substitute the values from c and e,

a = `(2sqrt(6))/((2sqrt(6))/(7))`

and

a = 7

a2= 49

Then, using the formula

b = `sqrt(a^2-c^2)`

or

b2 = a2 - c2

and substituting for a2 and c2,

b2 = 49 - (`2sqrt(6)` )2

b2= 49 - (4)(6)

b2= 49 - 24

gives the final required value of

b2 = 25

Then the equation of the ellipse is

`(x^2)/(49) + (y^2)/(25) ` = 1

Example parabola problem :

Reduce the equation y2-6y = 8x – 1 to standard form.

Solution:

Rearrange the equation of the 2nd -degree term and any 1st -degree terms of the same unknown are on the left side. Then group the unknown term appearing only in the 1st degree and all constants on the right:

y2 - 6y = 8x -1


Then complete the square in y:

y2 - 6y + 9 = 8x – 1 + 9

(y - 3)2 = 8x + 8

To get the equation in the form

(y - k)2 = 4a(x-h)

Factor an 8 out of the right side. Thus,

(y - 3)2= 8(x + 1)

is the equation of the parabola with its vertex at (-1,3).

Example hyperbola problem :

Find the equation of the hyperbola with an eccentricity of 3/2, directories x = ± 4/3, and foci at ( ± 3,0).

Solution:

The foci of the X axis at the points are (3,0) and ( - 3,0), so the equation is of the form ` (x^2)/(a^2)+(y^2)/(b^2)` = 1

First, the foci are given as (± 3,0); and since the foci are also the points (± c,0), then c=3

The eccentricity is given and the value of a2 can be determined from the formula

c = a * e

a = c/(e)

a = `(3)/((3)/(2))`

a =` (6)/(3)`

a = 2

a2 = 4

The relationship of hyperbola a, b and c are b2 = c2 – a2

And

b2 = (3)2 – (2)2

b2 = 9 – 4

b2 = 5

When these values are substituted in the equation `(x^2)/(a^2)+(y^2)/(b^2)` = 1 the equation `(x^2)/(4)+(y^2)/(5)` = 1

results and is the equation of the hyperbola.

The directories are given as x = `(4)/(3)` and, since

d = `(a)/(e)` or

a=de

Substituting the values given for d and e results in

a = `(4)/(3)((3)/(2))`

therefore,

a=2

and

a2 = 4

The value of c can be determined by the given equation,

c = ae

and a has been found to equal 2 and e is given as(3)/(2); therefore,

c =` 2((3)/(2))`

With the values of a and c computed, the value of b is found as before and the equation can be written.