Logarithm

In mathematics, the logarithm for a given number for a given base is the power or an exponent in which the base should be raised so as to create that number. For example, the logarithm or log of 10000 to base 10 is 4, because 4 is the power in which ten must be raised to produce 1000: 104 = 1000..

The logarithm of x to the base b can be written as log b(x) or, if the base is implicit, as log(x). So, for a number x, a base b and an exponent y,

If x = b y, then y = log b(x)
Laws of Logarithms:

While logarithm is just an exponent which is simply being written down on the following line.

Laws of logarithms
Example and Practice Problems on Logarithms to the Base 10:

Example 1:

Find the logarithm of 568 to base 10. Write this in the exponential form.

Solution:

Log 568 = 2.754

This indicates that102.754 = 5623

Mind investigation: 102 = 100 and 103 = 1000.

Our number 568 is between these two values, so it is reasonable.

Example 2:

Find the logarithm of 999 to base 10. Write this in the exponential form.

Solution:

Log 999 = 2.9999

This indicates that102.9999 = 5623

Mind investigation: 102 = 100 and 103 = 1000.

Our number 999 is between these two values, so it is reasonable.

My forthcoming post is on Markov Chains and Curve Fitting will give you more understanding about Algebra.

Example 3:

Find the logarithm of 8778 to base 10. Write this in the exponential form.

Solution:

Log 8778 = 3.943

This indicates that103.943 = 5623

Mind investigation: 103 = 1000 and 104 = 10000.

Our number 568 is between these two values, so it is reasonable.

Example 4:

Find the antilogarithm of the value -6.978

Solution:

This indicates that "if log N= -6.978, what is N?"

Using the laws of logarithm N = 10-6.978 = 0.000 000 105.

Practice problems:

Problem 1:

Find the logarithm of 4545 to base 10. Write this in the exponential form.

Solution:

The answer is 3.657.

Problem 2:

Find the logarithm of  555555 to base 10. Write this in the exponential form.

Solution:

The answer is 5.744.

Is this topic physics problems hard for you? Watch out for my coming posts.

Problem 3:

Find the logarithm of 8969 to base 10. Write this in the exponential form.

Solution:

The answer is 3.952.

Partial Fraction and Rules

A rational fraction is one which is a ratio of two polynomials.  That is, a function of the form

$f(x) = \frac{P_m(x)}{Q_n(x)}$

where $P_m(x) = a_mx^m + a_{m-1}x^{m - 1} + \ldots + a_2x^2 + a_1x + a_0$ is a polynomial of degree $m$, and $Q_n(x) = b_nx^n  + b_{n-1} x^{n-1} + \ldots + b_2x^2 + b_1 x + b_0$ is a polynomial of degree $n$.  Here, $a_m, a_{m-1}, \ldots, a_2, a_1, b_m, b_{m-1}, \ldots, b_2, b_1, b_0$ are constants, and $a_m \not=0, b_n \not=0$.

Depending upon the degrees of the numerator and the denominator, we can categorize these rational fractions into two types :

The fraction $f(x)$ is said to be an improper fraction, if $m < n$.  For example,

$\frac{3x^3 + 4x^2 - 11}{x^2 + 3x -2}$

The fraction $f(x)$ is said to be a proper fraction, if $m < n$.  For example,

$\frac{3x +4}{x^2 - 7x + 1}$

If $f(x) = \frac{P_m(x)}{Q_n(x)}$ is an improper fraction (i.e., $m\gen), then it can be expressed as a sum of a polynomial and a proper fraction.  That is,

$\frac{P_m(x)}{Q_n(x)} = A_{m-n} (x) +  \frac{R}{Q_n(x)}$

where, $A_{m-n}(x)$ is an polynomial of degree $m -n$, and $R$ is another polynomial of degree less than $n$.  For illustration, consider the following improper fraction:

$\frac{2x^3 + 4x^2 -11}{x^2 + 3x + 2} $ $= (2x -2) +$ $\frac{2x - 7}{x^2 + 3x + 2}$

Note that we can perform addition and subtraction on a group of rational fractions to get a rational fraction whose denominator is the lowest common denominator of the given fractions.  But the converse process of separating a fraction into a group of simple, or partial fractions  is often required.  It is natural to inquire about how far the decomposition of a given rational fraction into partial fractions.

Since we can always, by ordinary division, represent an improper fraction as the sum of an integral function and a proper fraction, we need to consider the resolution of a proper fraction into partial fractions.

Remark:

If the denominator of a proper fraction can be factorized into two integral factors, which are algebraically prime to each other, then we can always decompose the fraction into sum of two proper fractions.

Three cases along with the method of solving a rational fraction into a sum of partial fractions is given hereunder:
Partial Fraction Rules 1:

Case 1:If

$f(x) = \frac{A_m(x)}{B_n(x)} = \frac{A_m(x)}{b (x - \alpha_1)(x - \alpha_2) \ldots \ldots (x - \alpha_n)}$

is a proper rational fraction (i.e., m
$f(x) = \frac{A_1}{x- \alpha_1} + \frac{A_2}{x - \alpha_2} + \ldots \ldots + \frac{A_n}{x - \alpha_n}$

where $A_1, A_2, \ldots \ldots, A_n$ are constants.

Illustrative Example 1: Resolving $\frac{8x-4}{3x^2 - 2x -1}$ into a sum of partial fractions.
Solution

We know that

$3x^2 - 2x -1 = 3x^2 - 3x + x -1$

$=(3x^2 - 3x) + (x -1)$

$= 3x(x -1) + (x -1)$

$= (3x + 1)(x -1)$

Thus,

$\frac{8x -4}{3x^2 - 2x -1} = \frac{8x -4}{(3x+1)(x - 1)}$

Let us assume that

$\frac{8x -4}{(3x+1)(x-1)} = \frac{A}{3x+1} + \frac{B}{x-1}$

$\Longrightarrow \quad \frac{8x -4}{(3x+1)(x-1)} = \frac{A(x-1) + B(3x + 1)}{(3x+1)(x -1)}$

$\Longrightarrow \quad \frac{8x -4}{(3x+1)(x - 1)} = \frac{(A + 3B)x + (B - A)}{(3x + 1)(x -1)}$

Since the denominators are equal, we can equate the numerators.  Thus, we have

$8x - 4 = (A + 3B)x + (B - A)$

I am planning to write more post on List of Fibonacci Numbers and What is a Fraction. Keep checking my blog.

Comparing x coefficients and constants on both sides, we have

$A + 3B = 8$ ---------------(1)

$A - B = 4$ ----------------(2)

Solving (1) and (2), we have

$A =  5$ and $B = 1$.

Thus,

$\frac{8x -4}{3x^2 - 2x -1} = \frac{5}{3x+1} + \frac{1}{x -1}$.
Partial Fraction Rules 2:

Case 2:

When some of the zeros are repeating, we can write

$f(x) = \frac{A_m(x)}{b (x - \alpha_1)^{k_1} (x - \alpha_2)^{k_2} \ldots \ldots (x - \alpha_r)^{k_r}}$

where $r \leq n$; and $k_1, k_2, \ldots \ldots, k_r$ are positive integers such that $k_1 + k_2 + \ldots + k_r = n$ and $\alpha_1, \alpha_2, \ldots \ldots, \alpha_r$ are distinct zeros of $B_n(x)$.  In this case, the decomposition takes the following form:

$f(x) = \frac{A_1}{x - \alpha_1} + \frac{A_2}{(x-\alpha_1)^2} + \ldots + \frac{A_{k_1}}{(x-\alpha_1)^{k_1}} + \frac{A_{k_1 +1}}{x - \alpha_2} + \ldots  + \frac{A_{n - k_r +1}}{x - \alpha_r}+ \ldots + \frac{A_n}{(x - \alpha_r)^{k_r}}$

Illustrative Example 2: Resolving $\frac{3x^2 + x -2}{(x-2)^2 (1-2x)}$ into a sum of partial fractions.
Solution

Assume that

$\frac{3x^2 + x -2}{(x-2)^2 (1-2x)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{1-2x}$

$\Longrightarrow \quad \frac{3x^2 + x -2}{(x-2)^2 (1-2x)} =\frac{A(x-2)(1-2x) + B(1-2x) + C(x-2)^2}{(x-2)^2 (1-2x)}$

Since the denominators are equal, we can equate the numerators.  Thus, we have

$3x^2 + x -2 = A (x-2x^2 -2 + 4x)+ B(1-2x) + C(x^2 - 4x + 4)$

$\Longrightarrow \quad 3x^2 + x -2 = (-2A + C) x^2 +(5A - 2B -4C) x -(2A - B-4C)$

Comparing on both sides, we  have

$-2A + C = 3   -------------------- (1)$

$5A - 2B - 4C = 1    --------------(2)$

$2A -B-4C = 2    ------------------(3)$

From (1), (2) and (3), we get

$ A = - \frac{5}{3},  B = -4,  and C = -\frac{1}{3}$

Thus, we have

$\frac{3x^3 + x -2}{(x-2)^2 (1 - 2x)} = - \frac{5}{3(x-2)} -\frac{4}{(x-2)^2} - \frac{1}{3(1-2x)}$.
Partial Fraction Rules 3:

When some of the roots of the denominator $B_n(x)$ of the proper rational fraction $f(x) = \frac{P_m(x)}{Q_n(x)}$ are complex, we can no longer factor $B_n(x)$ in the form of $b(x-\alpha_1)^{k_1} (x - \alpha_2)^{k_2} \ldots (x - \alpha_r)^{k_r}$ with real factors.  However, if the coefficients are real numbers, then the complex roots always come in conjuage pairs of the form $\alpha \pm \beta i$ and that, corresponding to each such a pair of complex conjugate zeros, there is a quadratic factor of $B_n(x)$.

Illustrative Example 3: Resolving $\frac{2x+1}{(x -1)(x^2+1)}$ into a sum of partial fractions.

Assume that

$\frac{2x+1}{(x -1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

$\Longrightarrow \quad \frac{2x+1}{(x -1)(x^2+1)} =\frac{A(x^2 + 1) + (Bx + C)(x-1)}{(x -1)(x^2+1)}$

Since the denominators equal, we can equate the numerators.  Thus, we have

$2x+1= A(x^2 + 1) + (Bx + C)(x-1)$

$\Longrightarrow \quad 2x + 1 = (A+B)x^2 + (C- B)x + (A -C)$

Comparing on both sides, we have

$A + B = 0  -------------- (1)$

$C - B = 2 --------------- (2)$

$A - C = 1 --------------- (3)$

From (1), (2), (3), we have

$A = \frac{3}{2},  B = - \frac{3}{2}$, and $C = \frac{1}{2}$.

Thus, we have

$\frac{2x+1}{(x -1)(x^2+1)} = \frac{3}{2(x-1)} - \frac{3x-1}{2(x^2+1)}$

Numerical Variable

In mathematics the conception of a variable express the one quantity (the argument of the function also known as input) that determines the output. A function gives a unique value to each input of a specified type. A symbol is represented by a number is called variable. It is also called as numerical variable. We use letters for variables such as x, y, z, n, t.
Example for Numerical Variable:

Problem for numerical variable:1

Solve 5x + 3x = 4

Solution:

Given that 5x + 3x = 4

Add the two variables 5x and 3x is 5x + 3x = 8x

8x = 4

Take the 8 into after the equals

X = `4 / 8`

Divide the minimize the fraction `4 / 8` = `1 / 2`

So the answer will be x = `1 / 2` .

Problem for numerical variable: 2

Solve 19x – x + 2x = 19

Solution:

Given that 19x – x + 2x = 19

Subtract the 19x – x = 18x

18x + 2x = 19

20x = 19

Take the 20 to after the equals that will become `19 / 20`

x = `19 / 20` .

The answer will be   `19 / 20` .
Example for Numerical Variable:

I am planning to write more post on fourier transform, and real numbers. Keep checking my blog.

Problem for numerical variable:3

Solve 15x + 21x – 2x – 5 + 18 – x = 0

Solution:

15x + 21x – 2x – 5 + 18 – x = 0

Add the positive and negative symbols from the given equation

36x – 3x – 5 + 18 = 0

Subtract the variable 36x – 3x = 33x

Subtract the numerals -5 + 18 = 13

The equation will be 33x = 13

Take the 33 to after the equals x = 13 / 33

The solution will be x = 13 / 33

Problem for numerical variable :4

Solve x + 2x + x – 2x – 11 = 0

Solution:

Given that x + 2x + x – 2x – 11 = 0

Add the given variables x + 2x + x = 4x

4x – 2x – 11 = 0

Subtract the variable 4x -2x = 2x

2x – 11 = 0

Take the -11 into after the equal that will become +11

2x = 11

Take the 2 into after the equal

x = `11 / 2`

The answer will be 5.5.

Problem for numerical variable 5:

Solve -15x – 3x +2x – 4 = 0

Solution:

Given that -15x – 3x +2x – 4 = 0

Add the given variables – 15x – 3x = - 18x

- 18x + 2x – 4 = 0

Subtract the variable -18x + 2x = - 16x

- 16x – 4 = 0

Take the -4 to after the equals

-16x = 4

Take the – 16 to after the equals

-x = `4 / 16`

Minimize the fraction  - `4 / 16` to – `1 / 4`

x = – `1 / 4`

The answer will be – `1/ 4` .

List of Real Numbers

Let us study about list of real numbers. The various kinds of numbers together form the math problems for different methods.
Real numbers are also seemed to be as one type of numbers of math world. These real numbers are defined as the numbers that are with the combination of both definition of rational and irrational numbers.
Some of the examples for the list of real numbers are discussed in detail as below.

List of Real Numbers:

List of real numbers – example 1:

Add the following real rational numbers `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`

Solution:

The real rational numbers are `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`
Adding the given real rational numbers as follows:
`1/2 + 3/1 + 36/6 + 7/2 + sqrt(4) + 5.2`
0.5 + 3 + 6 + 3.5 + 2 + 5.2
Therefore the total value of the given real rational numbers is found to be as ’20.2’

List of real numbers – example 2:

Add the following real irrational numbers 23.232…, `sqrt(3)` , 3.3432…, `sqrt(7) and sqrt(5)`
Solution:

The real irrational numbers are 23.232…, `sqrt(3)` , 3.3432…, `sqrt(7) and sqrt(5)`
Adding the given real irrational numbers as follows:
23.232.. + `sqrt(3) ` + 3.3432.. + `sqrt(7) + sqrt(5)`
23.232.. + 3.3432..+ `sqrt(3) + sqrt(7) + sqrt(5)`
23.232.. + 3.3432.. + `sqrt(15)` (first adding all the square root values together)
23.232.. + 3.3432.. + 3.8729..
Therefore the total value of the given real irrational numbers is found to be as ’30.4481..’

List of real numbers – example 3:

Multiply the following real rational numbers `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`

Solution:

The real rational numbers are `1/2, 3/1, 36/6, 7/2, sqrt(4) and 5.2`
Multiplying the given real rational numbers as follows:
`1/2 * 3/1 * 36/6 * 7/2 * sqrt(4) ` `* 5.2`
0.5 * 3 * 6 * 3.5 * 2 * 5.2
Therefore the multiplied value of the given real rational numbers is found to be as ’327.6’

List of real numbers – exercises:

Add the following real rational numbers `12/3, 4/1, 18/6, 9/2, sqrt(36)` and 0.1 (Answer: 21.6)
Multiply the following real rational numbers `12/3, 4/1, 18/6, 9/2, sqrt(36)` and 0.1 (Answer: 129.6)
Add the following real irrational numbers 3.8782…, `sqrt(11)` , 0.372…, `sqrt(17)` and `sqrt(2)` (Answer: 9.3554…)

Simple Algebraic Expressions

Expressions are central concept in the algebra. A variable can take several values. Its values are not fixed. Otherwise, a constant has a fixed value. We are combining variables and constants to make algebraic expressions.

Operations of Simple algebraic expressions:

  For this, we can use the operations of addition, subtraction, multiplication and division. We have already come across simple algebraic expressions like 7x 10, 4y-3, 5+2zand so on. The above expressions were obtained by combining variables and constants. The expressions are also obtained by combining variables with themselves.
Simple Algebraic Expressions Example Problems:

1. Find the coefficients of y in the following expressions?

    4x – 3y –Coefficient of y is -3
     8 + yz, –Coefficient of y is z
    my + m- –Coefficient of y is m

2. State with reasons, which of the following pairs of terms are of like terms and which are of unlike terms:

    – 4ab, 7ba- Like terms Here ab=ba
    3xy, 3 x- Unlike Terms. The variable y is only in one term

Algebraic Expressions Practice Problems:

1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.

    Subtraction of z from y.
    One-half of the sum of numbers x and y.
    The number z multiplied by itself.
    One-fourth of the product of numbers p and q.
    Numbers x and y both are squared and added.
    Number 5 added to three times the product of numbers m and n.

Algebra is widely used in day to day activities watch out for my forthcoming posts on physics problems solutions and practice pre algebra. I am sure they will be helpful.

2.  Identify terms which contain x and give the coefficient of x.

(i) y2x + y        (ii) 13y2 – 8yx (iii) x + y + 2

(iv) 5 + z + zx  (v) 1 + x + xy  (vi) 12xy2 + 25

3. State whether a given pair of terms is of like or unlike terms.

(i) 1, 100          (ii) –7x, 5/2 x      (iii) – 29x, – 29y

(iv) 14xy, 42yx (v) 4m2p, 4mp2 (vi) 12xz, 12x2z2

Types Of Inequalities Geometry

An inequality is a statement about the relative size or order of two objects, or about whether they are the same or not (source: WIKIPEDIA). These inequalities will allow us to formulate several important geometric relationships. The types of inequalities geometry are transitive property, Substitution postulate, and trichotomy postulate. All the types of inequalities are used for proving inequality geometry.

Types of Inequalities Geometry
In geomeric inequality:
          A whole is equal to the sum of all its parts. A whole is greater than any of its parts.
         In geometry: The lengths of line segments and the measures of angle are positive number.
Consider these two applications:
                             
   If ACB is a line segment, then AB = AC + CB, AB > AC, and AB > CB.
   If ?DEF and ?FEG are adjacent angles, m?DEG = m?DEF + m?FEG,

m?DEG > m?DEF, and m?DEG > m?FEG                    
Types of inequalities:
Type 1: Transitive property of inequality:
      if a,b, and c are real numbers such that a > b and b > c, then a > c.
Then in geometry: If BA > BD and BD > BC, then BA > BC. Also, if m?BCA > m?BCD > m?BAC, then

m?BCA > m?BAC.
Type 2: Substitution postulate of inequality:
    A quantity may be substituted for its equal in any statement of inequality.
In geometry: If AB . BC and BC = AC, then AB > AC. Also, if m?C > m?A and m?A = m?B, then m?C > m?B.
Type 3: The trichotomy postulate Inequality:
Given  any two quantities, a and b, one and only one of the following is true;
                         a < b   or  a = b     or a > b.
Proving Inequalities and its Types in Geometry

My forthcoming post is on algebra math problem solver, math online solver will give you more understanding about Algebra.

Given: m?DAC = m?DAB + m?BAC and m?DAB > m?ABC

Prove: m?DAC > m?ABC
        Statements                                                               Reasons
1. m?DAC = m?DAB + m?BAC                                         1. Given.
2. m?DAC > m?DAB                                                       2. A whole is greater than any of its parts.
3. m?DAB > m?ABC                                                       3. Given.
4. m?DAC > m?ABC                                                       4. Transitive property of inequality.
This proof shows the inequalities geometry.