Limits of Logarithmic Functions

Introductions to Limits of Logarithmic Functions :
In mathematics, Logarithm is abbreviations of log functions. The logarithmic function contains three parts such as number, base, the logarithmic itself. This limits of logarithmic functions to discover logarithmic functions and their properties, such as domain, range, x and y intercepts and vertical asymptote. This function is called as the limits of logarithmic function. In this article we shall discuss about limits of logarithmic functions.

Definitions:

The logarithmic functions is called as inverse of exponential functions.

bN = A   is equivalent to N=logbA

Where ,        N = number functions,

A = logarithmic functions,

b = Base functions.

Properties for Limits of Logarithmic Functions:
1. Product functions:

`Log_x AB = log_xA+ log_xB`

2. Quotient functions   

`log_x(A/B)= log_xA - log_xB`

3. Power functions           

.`log_xA^B = B log_xA`

4.`LogBA xx logAB=1`

5.`Log_10(1)=0`

6. `Log_BB=1`

7.  `A ^( log_xA)` =x

Limits of Exponential Functions:

`lim_(x->0^+)`    loga x = `-oo` .    if a > 1

`lim_(x->0^+)`  loga x =  `oo`          if a < 1

`lim_(x->oo)` loga x=  `oo`           if a > 1

Having problem with functions and linear equations and inequalities keep reading my upcoming posts, i will try to help you.

Limits of Logarithmic Functions - Problems:

Limits of logarithmic functions - problem 1:

Solve the given limit function `lim_(x->0)` `log(1 + x^3)/cosx` .

Solution:

Given limit function is  `lim_(x->0)` `log(1 + x^3)/cosx` .

= `lim_(x->0)` `log(1 + x^3)/cosx` .

Apply the limit values in the above logarithmic function,   So, we get

=  `log(1 + 0^3)/cos0`.

we know the value of    03   and   cos 0

03   =  0    and   cos 0 = 1

So,                            =  `log(1 + 0)/1`.

= log 1

=  1 .

Answer:   1 .                              

Limits of logarithmic functions - problem 2:     

Solve the given limit function `lim_(x->0)` `log(5 - x^5)cosx` .

Solution:

Given limit function is  `lim_(x->0)` `log(5 - x^5)cosx` .

= `lim_(x->0)` `log(5 - x^5)cosx` .

Apply the limit values in the above logarithmic function,   So, we get

=  `log(5 + 0^5)cos0`.

we know the value of    05   and   cos 0

05   =  0    and   cos 0 = 1

So,                            =  `log(5 + 0)1`.

= log 5

Answer:   0.6989 .                             

Parametric Equations Calculus

Introduction for parametric equations calculus :

In mathematics, parametric equation is a method of defining a relation using parameters. A simple kinematical examples are  when one use a time parameter to determine the position, velocity, and other information about a body in motion. Abstractly, a Parametric Equations define a relation as a set of equation. It is therefore somewhat more accurates defined as a parametric representation. It is part of regular parametric equation representation.Calculus includes that differential calculus and integral calculus is used.(Source.Wikipedia)



Examples for Parametric Equations Calculus:

Example 1 : Prove that the sum of the intercept on the co-ordinate axes of any tangent to the curve x = d cos^4c,  y = d sin4c,   0 = ? =p /2   is equal to d.

Solution :

Take any point ‘C’ as (d cos^4c, d sin^4c, )

Now  `dx/dd ` = – 4d cos^3c sin c ;

And  `dx/dd` = 4d sin3c cos c

? `dy/dx` = –sin^2c/cos2c

Slope of the tangent at ‘c’ is = –sin^2c/cos2c

Equation of the tangent at ‘c’ is (y - d sin4c) =- sin^2c/cos2c(x - d cos^4c)

or x sin^2 c + y cos2 c = d sin^2 c cos2 c

?`x/d` cos2c+`y/d ` sin^2c= 1

sum of the intercepts = x cos2 c + y sin^2 c = d

Example 2 : Find the equations of the tangent and normal at B =p/2 to the curve x = b (B + sin B), y = b (1 + cos B).
Solution : We have

`dx/dB` = b (1 + cosB) = 2b cos2 B

2`dy/dB` = – b sin B = – 2b sin`B/2 ` cos`B/2`

Then dy/dx =`dy/dA` / `dx/dA` =  – tan`B/2`

Slope m = `dy/dx` [B = p/2] = – tanp/4 = –1

Also for B =p/2 , the point on the curve is{ b p/2 + b, b .}

Hence the equation of the tangent at B =p/2 is

y – b = (–1) [x – a(p/2 + 1)]

x + y =`1/2` b p + 2b or x + y –`1/2` b p – 2b = 0

Equation of the normal at this point is

y – b = (1) x – b(p/2+ 1)

x – y –`1/2` b p = 0

Practice Problem for Calculus Parametric Equations:

Find the equations of the tangents and normal to the ellipse x = c cosS, y = d sin S at the point S =p/4

Answer: (cx – dy) `sqrt(2)` – (c2 – d2) = 0.

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
I am planning to write more post on Definite Integrals, Newton Raphson Method. Keep checking my blog.
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.

I am planning to write more post on double digit multiplication, chat free online. Keep checking my blog.

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

Algebra is widely used in day to day activities watch out for my forthcoming post on algebraic expressions  I am sure they will be helpful.

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