Solving Equations with Radicals and Exponents

Solving equations with radicals and exponents:

Exponents:      

The term exponent in math is used to find the exponential value of the particular value it may be integer  or fraction . We can easy to get  the power of a  number using online calculator ,Consider  the unknown number , Here x is base and y is the power  of x . Let us consider the  number 23 Here 2 is the base(x)  and 3 is the power of 2(y). We calculate  23= 2 x 2 x 2 =8

Radicals:

Square root of the number is said to be a radical number .A radical equation is an equation in which a variable is under a radical term.Let us discuss about the solving an equations with radical and  exponents,

Example Problems to Solving Equations with Radicals and Exponents:

Example 1: Solve `sqrt(3x^2 +8x)` -3 =0

Solution:

Given `sqrt(3x^2+8x)` -3=0

isolate the radical equation,

`sqrt(3x^2 +8x)=3`

Take square on both sides we get,

`(sqrt3x^2+8x)^2` `=3^2`

3x^2+8x=9

3x^2+8x-9 =0

now you can solve the above equation using factoring method,

Using quadratic formula ,

x= `(-b+-sqrt(b^2-4ac))/(2a)`

Here b=8 ,a=3 and c=-9

Therefore ,

x= `(-8+-sqrt(8^2-(4)(3)(-9)))/(2(3))`

`=(-8+-sqrt(64+108))/6`

`=(-8+-sqrt(172))/6`

`=(-8+-sqrt(2*2*43))/6`

`=(-8+-2sqrt(43))/6`

`=2([-4+-sqrt(43)])/6`

`=(-4+-sqrt43)/3`

Therefore,

The factors are , `(-4+sqrt43)/3` ,`(-4-sqrt43)/3`

Example 2 : Solve `sqrt( x^2 +5x)-3=0`

Solution:

Isolate the given radical and exponent equation, we get,

`sqrt(x^2+5x)=3`

Take square root on both sides we get,

x^2+5x=32

x^2+5x=9

x^2+5x-9=0

Above equation in the form of ax^2+bx+c.

Therefore  you can find the factors using quadratic formula,

`x= (-b+-sqrt(b^2-4ac))/(2a)`

Here , b=5,a=1 and c=-9

substitute these values into formula,

`x= ((-5+-sqrt(5^2-4(1)(-9)))/(2(1)))`

`=(-5+-sqrt(25+36))/2`

`=(-5+-sqrt(61))/2`

`=(-5+-sqrt(61))/2`

Therefore ,

The factors are ,

`x= (-5+sqrt61)/2`    ,   `(-5-sqrt61)/2`

More about the Solving of Equations with Radicals and Exponents:

Example3: Solve `sqrt(2x^2+4)=4`

Solution:

Take square on both sides we get,

`(sqrt(2x^2+4))^2` =42

2x^2+4 =16

subtract both sides by 4,we get

2x^2+4-4=16-4

2x^2=12

Divide both sides by 2 ,

x^2 =`12/2`

x^2=6

x=`sqrt6`

Example 4: Solve the equations with radicals  and exponents `sqrt(4x^2+8)=11`

Solution:

Take square on both sides , we get,

4x^2+8 =121

Subtract both sides by 8 we get,

4x^2+8-8 =121-8

4x^2=113

Divide both side by 4,

x^2 =113/4

x=`sqrt(113/4)`

x=`(sqrt113)/2`

Therefore the value of `x= sqrt113/2`

Example 5: Solve the equation with radicals and exponents `sqrt(6x^2+7)` `=8`

Solution:

Take square on both sides we get,

6x^2 +7 = 64

Subtract both side by7,

6x^2+7-7=64-7

6x^2=57

Divide both side by 6,

x^2 =`57/6`

x= `sqrt(57/6)`

Therefore the value of `x = sqrt(57/6)`

Interest Compounded Quarterly

Introduction to interest compounded quarterly:

Interest is a fee paid on borrowed assets. It is the price paid for the use of borrowed money. Compound interest arises when interest is added to the principal, so that from that moment on, the interest that has been added also itself earns interest. This addition of interest to the principal is called compounding (for example the interest is compounded). In this article we shall discuss about interest compounded quarterly

Interest Formula for Compounded Quarterly

The basic formula for Compound Interest is:

FV = PV (1+r)n

PV is the current value or present value

r is the annual percentage rate of interest (percentage)

n is the total number of years the amount is deposit

FV = Future Value (amount of money collect after n number of years, with interest.)

Quarterly compounded interest = P (1 + r/n)nt = (Quarterly Compounding)


Interest Compounded Quarterly Example Problem

Ex 1:Rose deposits $7000 in a bank account, bank paying at the rate of 7% per year, compounded and credited quarterly. Find how much will he have at the end of 5 years?

Here p=$7000, n=4, r=7/100, t=5

Quarterly compounded interest = P (1 + r/4)4(5)

=7000(1+0.07/4)20

= 7000.00 is worth  9,903.45

Ex 2:Jessica deposits $6000 in a bank account, bank paying at the rate of 6% per year, compounded and credited quarterly. Find how much will he have at the end of 3 years?

Here p=$6000, n=4, r=6/100, t=3

Quarterly compounded interest = P (1 + r/4)4(3)

=6000(1+0.06/4)12

= 6000.00 is worth  7,173.71

Ex 3:Joseph deposits $5000 in a bank account, bank paying at the rate of 5% per year, compounded and credited quarterly. Find how much will he have at the end of 4 years?

Here p=$5000, n=4, r=5/100, t=4

Quarterly compounded interest = P (1 + r/4)4(4)

=5000(1+0.05/4)16

= 5000.00 is worth  6,099.45

Ex 4:Jim deposits $4000 in a bank account, bank paying at the rate of 5% per year, compounded and credited quarterly. Find how much will he have at the end of 6 years?

Here p=$4000, n=4, r=5/100, t=6

Quarterly compounded interest = P (1 + r/4)4(6)

=4000(1+0.05/4)24

= 4000.00 is worth  5,389.40

Domain of a Logarithmic Function

Introduction to Domain of a Logarithmic Function
In general, let us consider a function,  y = f(x).

It means, y the value of the function varies depending upon the input variable x and nature of the function.

As long as any value of x makes gives a real value of y, the function said to exist all the time. But this may not be the case with many functions due to the nature and restriction of the functions. In such a case, only for a particular set (or sets) of values of the input variable the function exists.

The set (or sets) of values of the input variable which makes the function exist is called the domain of the function and the corresponding set (or sets) of values of the function is called as the range of the function.

Let study the domain of a logarithmic function.

Description of a Domain of a Logarithmic Function

Before determining the domain of a logarithmic function, let us see what a logarithmic function is.

Mathematicians discovered that the function could be described in the form

f(x) = logbx, which is called as logarithmic function.

Let, n = logbx   Then as per the definition of a logarithmic function, bn = x

Determining the Domain of a Logarithmic Function

To determine the domain of a logarithmic function, let us start from the fundamental concept.

Let us take the simple form of a logarithmic function  y = logbx

Then as per definition   by = x
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If you carefully notice, the value of x becomes closer and closer to 0 as y becomes infinitely smaller and smaller. As ultimate, only when y becomes – infinity, the value of x becomes 0, which is an impossible situation. The following graph shows the variation.

Reversing the above argument, it can be said that a logarithmic function exists only for all values greater than 0.

That is domain of a logarithmic function is x > 0

Please note that for simplicity we assumed f(x) = logbx. In general it could be a logarithmic function of another function of x.

That is, f(x) = logb g(x).

In such a case, the domain of the logarithmic function is determined by solving  g(x) > 0

For example,  if f(x) = logb (x – 1), then the domain of the function is x > 1.

Fraction Decimal Notation

Introduction to fraction decimal notation:

Let us discuss the fraction decimal notation. The fraction notation is defined the part of whole number. The fraction is generally specifying two parts. The first part is numerator and second part is denominator. The numerator part is top part and denominator is bottom part. The decimal notation is same as the fraction notation. Decimal notation has decimal point. The example is 3.16.

Fraction Decimal Notation:

The fraction notation is numerator / denominator. Example is 4 / 6. The top number is known as the numerator and bottom number is known as the denominator.

The 4 is called numerator.
The 6 is called the denominator part.
The example of the fraction number is 9 / 7, 5/ 6 and etc.

The decimal point is very important part of the decimal notation. The example of the decimal notation is 63.147.

The 63 is called  the whole number.
The dot (.) is called  the decimal point.
The 1 is called the tenth place of the number. The meaning is 1/10.
The 4 is called the hundredth place of the number. The meaning is 4/100.
The 7 is called the thousandth place of the number. The meaning is 7/1000.
The fraction notation is written the decimal notation. The example is 5/7 is fraction notation. The decimal notation of 5/7 is 0.71.

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Example Problem of Fraction Decimal Notation

Problem 1:

Divide the 8 by 10.

Solution:

The 8 is dividend and 10 is divisor

0.8
____
10  ) 8
0
------
80
80
------
0
-------

The 8/10 decimal value is 0.8.

Problem 2:

Divide the 3 by 2.

Solution:

The 3 is dividend and 2 is divisor

0.666
______
3  ) 2
0
------
20
18
------
20
18
-----
2
------

The 3/2 decimal value is 0.66.

Logarithm of Complex Number

Introduction to logarithm of complex number:


In this article we will study about logarithm of complex number. In this foundation is the higher grade mathematics. Logarithm is used to solve many complex problems in easy way. In this content we are going to discuss about logarithm of complex number. The following are the examples involved in logarithm of complex number

Sample Problem for Logarithm of Complex Number:

Logarithm of complex number Problem 1:

Solve the given logarithmic expression: in 3x + in 5 = 3

Solution:                                                                                                                            

Given logarithmic expression: ln 3x + ln 5 = 3

in 3x + 1.6094 = 3                   ( The value of ln 5 = 1.6094)

Subtract by 1.6094 on both side in the above expression.

ln 3x + 1.6094 - 1.6094 = 3 - 1.6094

ln 3x =  1.3906   

log e 3x = 1.3906                        (ln x = loge x)        

(We know, y = logbP       and P = by)

Here, y = 1.3906         P = 3x                b = e

So,    3x = e1.3906   = 4.0172

3x = 4.0172

Divided by 3 on both side in the above equation

=3x/3 = 4.0172/3

x = 1.3390

Answer: The value of x = 1.3390

Logarithm of complex number Problem 2:

Solve the problems `log 10^x = log^5`

Solution:

Step i: given  ` log 10^x = log^5`

Step ii:  simplify the equation using with law (when the log base to the power here we can written like multiplication of log term)

` x.log10=log^5`

Step iii: evaluate for x, here we can dividing each side by log10



`x = (log^5/log10)`

`x = (log^5/log10)`

Or

Step iv: Take log value for the terms

Log 5= 0.69900

Log 10=1

` x =0.69900/1`

Step v: x = 0.699.

Logarithm of complex number Problem 3:

Solve the problems 61 log29+61 log27 = 66 log2 (7x)

Solution:

Logarithmic function 61 log25+61 log27=61 log2 (7x)

61 log2(9*7)  =61 log2(7x)

log2(63)=log2(7x)  by logarithm rules

Equate both sides, s the base are same 2, 7x = 63

Simplification: `x =63/7`

Answer = 9

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Practice Problem for Logarithm of Complex Number:

Get the values of given problems `log^8 + log^4 + log^5`
Answer:log (160)

2. Solve the given logarithmic expression: In 6x + In 5 = 6

Answer: x = 0.163390

Easy Way to Learn Limits

Introduction to Easy way to learn limits:      
In the easy way to learn limits, let f be a function of a real variable x. Let c and l be two unchanging numbers. If f(x)  come within reach of the value l as x approaches c, we say l is the limit of the function f(x) as x tends to c. This is written as

`lim_(x->c)`   f(x)   =  l.

Left Hand and Right Hand Limits:

In thr easy way to learn limits, although defining the limit of a function as x be inclined to c, we consider values of f(x) when x is very close to c (x>c or x
Lf(c) = `lim_(x->c )` _f(x) , provided the limit exists.

Likewise if x gets merely values greater than c, next x is said to tend to c from above or from right, and is denoted symbolically as x ? c + 0 or x ? c+, here the limit of f is then labeled the right hand limit. This is written as Rf(c) = `lim_(x->c+)` f(x). It is significant to reminder that for the survival of`lim_(x->c)` f(x) it is necessary that both Lf(c) and Rf(c) exists and Lf(c) = Rf(c) =`lim_(x->c)` f(x).  As well as the left and right hand limits are labeled as one sided limits.

Fundamental Results for Easy Way to Learn Limits:
The following rules for easy way to learn limits,

(1) If f(x) = k for all x, then
`lim_(x->c) `  f(x) = k.
(2) If f(x) = x for all x, then                                                                                                               
`lim_(x->c)`  f(x) = c.
(3) If f and g are two

functions having limits and k is a invariable then

(i)  `lim_(x->c)` k f(x) = k `lim_(x->c)` f(x)

(ii)  `lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) + `lim_(x->c)`  g(x)

(iii)  `lim_(x->c)`  [f(x) - g(x)] = `lim_(x-> c)`  f(x) - `lim_(x->c)`  g(x)

(iv)  `lim_(x->c)` c [f(x) . g(x)] = `lim_(x->c)` f(x) . `lim_(x->c)` g(x)

(v)    `lim_(x->c)` [`f(x)/g(x)` ] = `lim_(x->c)`  f(x)  / `lim_(x->c)`  g(x)              g(x) `!=` 0

(vi)  If   f(x) = g(x) then

`lim_(x-> c)`  f(x) =  `lim_(x->c)` g(x).

Example of Easy Way to Learn Limits:

Evaluate  `lim_(x->3)` `(x^2+ 7x + 11) / (x^2-9)` .

Solution:
Let f(x) = `(x^2 + 7x + 11)/( x^2-9)`
This is of the form f(x) =`g(x) / (h(x))`  ,
where g(x) = x2 + 7x + 11 and h(x) = x2 - 9. Clearly g(3) = 41 ? 0 and h(3) = 0.
Therefore f(3) = `g(3) /(h(3))`  = `41/ 0`  . Hence `lim_(x->3)`  `(x^2+ 7x + 11) / (x^2-9)` does not exist.

Solving Plane Geometry

Introduction to solving plane geometry:

Geometry is one of the important parts of mathematics deals with properties of shape of geometric objects. Geometry used to say all kinds of shapes and their properties. There are two major classifications in geometry. They are Plane geometry and Solid geometry. Shapes that are drawn at flat surface are called Plane and the study about this is called  Plane geometry. Plane geometry is about the shapes of lines, circles, etc. Plane geometry is study about two-dimensional objects.

Problems on Solving Plane Geometry:

Problem 1:

Solving the area of the triangle with the base of 7cm and the height is 10cm.

Solution:

Given:       Base = 7cm

Height = 10cm

Area = (b * h) / 2                       

= (7 * 10) / 2                            

= 70 / 2

= 35 cm^2

Problem 2:

Solving the area of the circle with 10.5cm radius.

Solution:

Given:     Radius = 10.5cm

Area of the circle =`pi` r2

= 3.14 * 10.5 * 10.5

= 346.18cm^2

Problem 3:

Solving the area of the trapezoid whose length is 10 cm, the width is 5cm and the height is 10 cm.

Solution:

Given:   Length = 10cm

Width = 5cm

Height = 10cm

Area = h (l + w) / 2                                                                                                                

= 10 (10+5) / 2

= 150 / 2

= 75cm3

Problem 4:

Solving the area of the rectangle with length 6.5cm and width 3.5cm. Also find its perimeter.

Solution:

Given:    Length = 6.5cm

Width = 3.5cm

Area = L * W

= 6.5 * 3.5

= 22.75cm^2            

Perimeter = Sum of all sides

= 2L + 2W

= 2 * 6.5 + 2 * 3.5

= 13 + 7

= 20cm^2

Problem 5:

Solving the area and perimeter of the parallelogram whose Height 12 cm, length 14cm, width 16cm.

Solution:

Given:    Length = 14cm

Width = 16cm

Height = 12cm

Area = L * W

= 14 * 16

= 224cm^2

Perimeter = Sum of all sides

= L + W + H

= 14+16+12

= 42cm3


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Practice Problems on Solving Plane Geometry:

1.   Find the area of the trapezoid whose length is 3 cm, the width is 1.5cm and the height is 5cm.

Answer:  11.25cm^3

2.  Find the area of the circle with 0.5cm radius.

Answer:  0.785cm^2

3.  Calculate the area and perimeter of the rectangle whose length is 2cm, width is 2.5cm.

Answer:   Area = 5cm^2
Perimeter = 9cm^2

Distribution of Prime Numbers

Introduction of distribution of prime numbers:

The distribution of positive numbers is divided into various terms, such as prime numbers like 2, 3, 5, and 7. The distribution of prime numbers cannot be resolute into the factors and the composite numbers. The fundamental theorem of arithmetic gives the most importance for the distribution of prime numbers. Hence the composite numbers can be expressed in a way as the product of the prime numbers.

Distribution of Prime Numbers:

The problem on the threshold of math is the queries of the distribution of the prime numbers. Though the sequence of the prime numbers shows the irregularities with explanation, the distribution of the prime numbers is created to possess the samples on the regularities in some of the terms and to make mathematical investigation.

The p(x) can be denoted as the number of prime numbers that do not exceed the value x. The resolving of the function can be done in the study of the function p(x). The examination of the table of prime numbers at the observance however the table gives the prime numbers which shows no signs which coming to an end. They become to the average which widely spaced in the higher position of the table. Theorems which gives the distribution of the prime numbers. These theorems would results to p(x) at the infinity condition.

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Theorems in Distribution of Prime Numbers:

The initial theorems which shows the infinite number of the prime numbers. Let we have the P as the multiple of the finite sets of the prime numbers. Let we indicate Q = P + 1, the letters P and Q which don’t have any prime factors. These factors would divide Q – P = 1, this is not possible. Hence the Q may be the divisible of the prime number. Here exists of one prime number from occurring of P.