The Math Blog

Converting Logarithmic to Exponential Equation

2016-03-25T04:32:00.001-07:00

Logarithms are the opposite of exponents. Thus, a logarithmic equation can be converted to an equivalent exponential equation.

`log_2(8) = 3` is the same equation as `2^3 = 8`.

Any logarithmic equation has three parts:

Base of logarithm (the number written in subscript after logarithm; 2 in the above example)
Argument of logarithm (the number written after the base in line with 'log'; 8 in the above example)
Result (the number after the = sign; 8 in the above equation)

Similarly, any exponential equation has three parts:

Base of exponent (2 in the above example)
Exponent (or power, or index; 3 in the above example)
Result (8 in the above example)

To convert a logarithmic equation to an exponential equation, we use the following concepts:

Base of logarithm becomes base of exponent
Argument of logarithm becomes result of exponential equation
Result of logarithmic equation becomes exponent

Consider the logarithmic equation `log_{10}(100) = 2`,

Base is 10
Argument is 100
Result is 2

According to the concepts explained above, we will change the above numbers as follows:

Base of logarithm, 10, becomes base in the exponential equation
Argument 100 becomes result of exponential equation
Result of logarithmic equation, 2, becomes exponent

Thus, the equivalent exponential equation to `log_{10}(100) = 2` is `10^2 = 100`.

Difference between Sequence and Progression

2016-03-23T02:29:00.001-07:00

A sequence is a set of numbers in a specific order such that you can predict the next number by a specific logic. For example, the following set of prime numbers:

2, 3, 5, 7...

is a sequence, because

it is a set of numbers
separated by commas
in order of a specific logic (numbers are prime)

On the other hand, a progression is a sequence of numbers which has a specific mathematical formula. That is, you can predict the next term of a progression by the help of a formula, but you can't always have a formula to predict the next number of a sequence.

The set of prime numbers is a perfect example to differentiate between sequences and progressions; It is no doubt that the set of prime numbers is a sequence, but it is not a progression because there is no specific formula to predict the next prime number.

Converting Logarithms to Exponential Form

2016-03-20T10:36:00.001-07:00

Consider the following logarithm equation:

`log_{10}(100) = 2`

and the following exponential equation:

`10^2 =100`

We can say that both the above equations are equivalent. They are just different forms of the same equation.

Every logarithmic expression can be converted to an exponential expression by remembering the following rules:

The base of the logarithm is the base of the exponential expression
The number/quantity which the logarithmic expression equals is the exponent in the exponential expression
The argument of the logarithm is the result of the exponential expression

Thus, relating to the above two equations, we can write the following steps:

Base of the logarithm `log_{10}(100)` is 10, hence 10 will be the base of the exponential expression.
The logarithmic expression `log_{10}(100)` equals 2. Thus, 2 is the exponent. That is, `10^2`
The argument of the logarithmic expression is 100. This becomes the result of the exponential expression. Thus, we can write `10^2 = 100`.

Following the above three steps, you can convert any logarithmic expression/equation to its equivalent exponential expression/equation. A more detailed post on this topic is available here.

Solved Example

Convert the following logarithmic equation to exponential form:

`log_{2}(32) = 5`

First, identify the base, argument and result of the logarithmic equation:

Base = 2
Argument = 32
Result = 5

Now using the rules described above, we know the following things about the exponential expression:

The base 2 of the logarithm will become the base of the exponential expression
The argument 32 of the logarithm will become the result of the exponential expression
The result 5 of the logarithmic equation will become the exponent

Thus, the exponential equation equivalent to the above equation is,

`2^5 = 32`

Worksheet

Instructions

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

Answer

`log_2(3) = 8`

Show Answer

`log_{10}(1000) = 3`

Show Answer

`log_{10}(10) = 1`

Show Answer

`log_{5}(25) = 2`

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`ln(e) = 1`

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Evaluating Logarithms

2016-03-20T01:28:00.003-07:00

The value of a logarithm can be calculated by two methods:

Converting to Exponential Form
Using Logarithm Table

Converting to Exponential Form

As dicussed earlier in this post, a logarithmic expression is equivalent to an exponential expression since logarithms are the opposite of exponents.

Suppose we have the logarithmic expression

`log_{10}(100)`

Let the expression be equal to 'x'. (we have to eventually find the value of 'x'),

`log_{10}(100) = x`

Convert to exponential form,

`10^x = 100`

Write 100 as an exponent of 10. We have `100 = 10 * 10 = 10^2`. Thus,

`10^x = 10^2`

Equate the two expressions. Two exponential expressions are equal when two their bases and exponents are both equal. In the above equation, the bases are 10 in both the expressions and the exponents are `x` and 2. Thus `x` should be equal to 2. Thus,

`x = 2`

Now that we have the value of 'x', we will equate the original expression to 'x',

`log_{10}(100) = 2`

Solved Examples

Solved Example 1

Evaluate the following logarithms:

`log_{10}(1000)`

Let the expression be equal to 'x',

`log_{10}(1000) = x`

Convert to an exponential expression,

`10^x = 1000`

Write 1000 as a power of 10. We have `1000 = 10 * 10 * 10 = 10^3`. Thus,

`10^x = 10^3`

Equate the two exponential expressions. Their bases are both equal to 10. Their exponents are `x` and `3` respectively. Thus,

`x = 3`

Thus, we can write the original expression equal to 'x'.

`log_{10}(1000) = 3`

Worksheet

Using Logarithm Table

Using the logarithm table, we can find the value of logarithms with base 10 and natural base only. However, if you know the base changing formula of logarithms, you can convert a logarithm of any base to logarithms of base 10, evaluate them, and then find the value of the original logarithm with different base.

First we will discuss finding the value of a logarithm with base 10 using a logarithmic table. Suppose you have to evaluate the following logarithm

`log_{10}(100)`

Base and Argument of Logarithm

2016-03-16T02:33:00.002-07:00

What are Base and Argument in a Logarithm?

Consider the following expression:

`log_{10}(2)`

In the above expression,

The number 10 is called the base of the logarithm
The number 2 is called the argument of the logarithm

The above expression is read as

Logarithm of 2 to the base 10.

We can see that

The base is always written in subscript (below) just after the word 'log'
The argument is the number written in brackets after the base. It is written in line with the word 'log'

What do the base and argument mean in a logarithm?

If we refer to a logarithm table, the value of `log_{10}(2)` is 100. That is,

`log_{10}(2) = 100`

We already know that logarithms are the opposite of exponents. The above expression if converted to exponents is

`10^2 = 100`

In the above expression, we know that 10 is the base raised to the power (exponent) of 2. Thus we call 10 the base of the logarithm `log_{10}(2)`

The number 2 is called the argument because logarithm (to the base 10) is a function to which we pass the value '2' in order to get the result 100.

What if you change the base of the above logarithm?

Suppose we change the base of the above logarithm to 5. What will the result be then?

`log_5(2) = ?`

Let the result be 'x'.

`log_5(2) = x`

Since 5 is the base and 2 is the argument, we will raise 5 to the power of 2 to get 'x'.

`5^2 = x`

Evaluate `5^2 = 5 * 5 = 25`, thus,

`5^2 = 25`

Thus x is 25 and the logarithm's result is 25.

`log_5(2) = 25`

Thus, base of a logarithm is as important as its argument. Changing it changes the meaning and value of the logarithm.

Special Cases

If the base of a logarithm is not specified, we assume it to be 10. Thus, `log(2)` is same as `log_{10}(2)`
If the base of a logarithm is not specified and instead of 'log' there is 'ln' (which means natural logarithm) then the base is the exponential constant, `e`. Thus, `ln(10)` is same as `log_e(10)`.

Solved Examples

Identify the base and argument of the following logarithms:

`log_{10}(5)`
`log_{100}(2)`
`log_x(y)`
`log(10)`
`ln(100)`

Answers:

The number written in subscript after log is the base. Thus, 10 is the base. The argument is the number written in line with log after the base. Thus, 5 is the argument.
100 is the base and 2 is the argument.
The variable 'x' is written in subscript after log. Thus the base is `x`. `y` is written after the base and it is in line with 'log'
No base is specified. Thus the base is 10.
No base is specified and `ln` is written instead of `log`. Thus, the base is the exponential constant, `e`.

Worksheet

Instructions:

Find the base of the following logarithms

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

Answer

`log(3)`

Show Answer

`log_{100}(50)`

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`log_e(10)`

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`log_{a}b`

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`log_{2^3}(8)`

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Instructions:

Find the argument of the following logarithms:

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

Answer

`log(e)`

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`log(10)`

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`log_{10}(5)`

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`log_{n}(2)`

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`ln(2)`

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General Form of a Quadratic Expression

2016-03-11T20:30:00.000-08:00

A quadratic expression written in the following format is said to be in the general form.

`ax^2 + bx + c`

Characteristics of the general form:

`a`, `b` and `c` are numbers
`a` does not equal 0
`b` and `c` can be equal to 0

Description of the general form:

When a quadratic equation or any algebraic equation is written in descending order of exponents of its variable, it is said to be in the general form.

The above equation can be written as

`ax^2 + bx^1 + cx^0`

Notice that the first term in the general form has an exponent of '2' on `x`, the second term has an exponent of 1 on `x` and the third term has an exponent of 0 on `x`. Thus, in the general form, the terms are written in descending powers of `x`.

Examples

The following quadratic expressions are in the general form:

`x^2 + 2x + 1`
`10x^2 + 3x + 4`
`4x^2 - 8x + 4`

...while the following quadratic equations are not in the general form:

`x + 2x^2 + 5`
`1 + 3x + 3x^2`
`x^2 + 1 + 2x`

Forms of a Quadratic Expression

2016-03-11T20:05:00.001-08:00

A quadratic expression can be written in three different ways:

General Form

`ax^2 + bx + c`

Intercept Form

`a(x - p)(x - q)`

Vertex Form

`a(x - h)^2 + k`

Each of the above forms has its own significance, which is discussed in each form's separate post.

Worksheet: Identifying Quadratic Expressions

2016-03-11T19:59:00.000-08:00

Instructions

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

Answer

`x^2 + 3x + 4`

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`x^3 + 3x^2 + 1`

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`x + 1`

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`xy + 2x^2 + y^3`

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`xy^2 + x + 1`

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`x^2 + 2xy + y^2`

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`(x^3)/(x) + 2x + 1`

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Solved Examples: Identifying Quadratic Expressions

2016-03-11T19:36:00.001-08:00

Identify whether the following expressions are quadratic or not, and state why or why not they are quadratic:

`x^2 + 2x^3 + 3`

This expression is not quadratic because it contains a term with exponent greater than 2 on the variable `x`.

`x + 2`

This expression is not quadratic because it does not contain a term with exponent '2' on the variable.

`x^2 + 2x + 1`

This expression is quadratic because

the greatest exponent on `x` is 2
`x` does not have negative exponents or exponents greater than 2

`xy + x^2 + y^2`

The above expression is quadratic because its degree is 2. In other words, it is quadratic because the sum of exponents of `x` and `y` in any single term is never greater than 2.

`x^2y + xy + 1`

This expression is not quadratic because the sum of exponents of `x` and `y` in the first term is 3.

Identifying Quadratic Expressions

2016-03-11T03:11:00.000-08:00

Definition of Quadratic Expressions

A quadratic expression is an algebraic expression having a degree of 2.

Identifying One Variable Quadratic Expressions

An algebraic expression in one variable is quadratic when:

The variable has an exponent of 2 on it
The variable does not have an exponent greater than 2
The variable does not have a negative exponent

In simple words the above three statements mean:

There is at least one term in the expression with `x^2` in it
There is no term in the expression with `x^3` or higher exponents in it
The expression does not have `x` in the denominator

Example

The algebraic expression `x^2 + 2x + 3` is quadratic because

'x' has an exponent of '2' on it
'x' does not have an exponent greater than 2 on it, and
'x' has an exponent of 1 in the term '2x' and an exponent of 0 in the term '3'

Counter Examples

The expression `x^3 + 3x^2` is not quadratic because 'x' has an exponent greater than 2 in it.
The expression `x^2 + 1/x` is not quadratic because 'x' is in the denominator (and hence has negative exponents)
The expression `0x^2 + 3x + 2` is not quadratic because there is no term with `x^2` in it (the term `0x^2` equals 0 so it is not considered)

Identifying Two Variable Quadratic Expressions

A quadratic expression can have two variables, given that

the sum of exponents of both variables in any term is never greater than 2, and
there is at least one term whose degree is 2
no variable is present with a negative exponent

Example

`x^2 + xy + y^2` is a quadratic expression because

`x` and `y` do not have exponents greater than 2 in any term
sum of exponents of x and y in the term `xy` is 2

Counter Examples

`x^2y + xy + 1` is not a quadratic expression because the sum of exponents of `x` and `y` in the term `x^2y` is 3, which is greater than 2
`x/y + x^2` is not a quadratic expression because `y` is in the denominator, thus making it a rational expression

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Identifying Quadratic Equations Next:
Solved Examples: Identifying Quadratic Expressions

Quadratic Expressions

2016-03-09T16:48:00.000-08:00

An algebraic expression having the degree of '2' is called a quadratic expression. By 'degree', we mean the highest exponent on the variable/s in an algebraic expression.

For example, `x^2 + 2x + 3` is an algebraic expression in which the variable `x` has the highest exponent of `2`.

Quadratic expressions can be written in various forms, namely the General Form, the Vertex Form and the Intercept Form. A quadratic expression can be converted from one form to another by using algebraic operations.

A quadratic expression can be factored into a product of two linear expressions by the method of splitting the middle term.

Previous:
Introduction to Quadratic Equations Next:
Identifying Quadratic Equations

Introduction to Quadratics

2016-03-09T03:45:00.000-08:00

Quadratics is the branch of mathematics which deals with factoring quadratic expressions, solving quadratics equations and graphing quadratic functions.

A quadratic expression is an algebraic expression having the degree of 2. It is typically an expression in the variable 'x' (or any other variable) having the highest exponent of 2. An example is

`x^2 + 2x+ 3`

A quadratic equation is an equation in which we equate a quadratic expression with zero. Thus, the following is a quadratic equation:

`x^2 + 2x + 3 = 0`

A quadratic equation does not have more than two solutions. They are called the roots, zeros or x-intercepts of the quadratic equation.

A quadratic function is the equation of two variables in which the dependent variable is equal to a quadratic expression in the independent variable. For example, in the following equation:

`y = x^2 + 2x + 3`

the dependent variable is 'y' and it is equal to a quadratic expression in the independent variable 'x'.

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Quadratics Home Page Next:
Quadratic Expressions

Convert quadratic function from standard to vertex form

2016-03-04T02:18:00.001-08:00

The standard form of a quadratic equation is as follows:

`ax^2 + bx + c = 0`

It's vertex form is as follows:

`a(x- h)^2 + k = 0`

A quadratic equation can be written in both forms. It can be converted from the standard to the vertex form by the method of completing the square.

The following example explains the method.

Consider the quadratic equation in general (standard) form:

`x^2 + 2x + 3 = 0`

We will convert it to vertex form by the following steps:

Move the constant term to the right hand side

The constant term (number without a variable) in the above equation is '3'. We can move it to the right hand side of the = sign as follows:
`x^2 + 2x = -3`

Identify the coefficient of 'x'

The coefficient (number near) 'x' in the above equation is '2'.

Divide the number by 2

Dividing the number '2' by 2, we get 1.

Complete the square

Notice that `2x` in the equation has a `+` sign before it. Thus, we will take the number obtained in the previous step and add it with 'x' and then square the expression as follows:

`(x + 1)^2`

Now we can replace the left hand side of the quadratic equation with the above expression as follows:

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

Move the constant term in the right hand side again to the left hand side of the equation

`(x + 1)^2 + 3 = 0`

The above equation is said to be the vertex form of `x^2 + 2x + 3 = 0`

Favorable Outcomes

2016-02-29T03:00:00.003-08:00

Favorable outcomes in probability are those results of an experiment which you want.

Favorable outcomes of a coin

Suppose two friends toss a coin. Friend A wants 'heads' and B wants 'tails'. We can say that the favorable outcome for A is 'heads' and for B is 'tails'.

When a coin is tossed, it can result in either heads or tails. Thus, there are two possible outcomes on tossing a coin. Any one of these outcomes, heads or tails, can be favorable according to what you want.

If suppose someone says they want either heads or tails, then the favorable outcomes will be 2, because whether the coin gives heads or tails the outcome is what they wanted.

Favorable outcomes of a number cube

A number cube has six faces numbered 1 through 6 and on throwing it only one of these faces shows up.

Suppose you have to calculate the probability of getting an even number on rolling a number cube. The number cube has the numbers 1, 2, 3, 4, 5 and 6 on it, out of which only 2, 4 and 6 are even numbers. Thus, there are a total of 6 possible outcomes but only 3 favorable outcomes. If any one of these favorable outcomes occurs, then you say that the result is favorable, or of interest.

Solved Examples: Difference of Cubes

2016-02-10T11:34:00.001-08:00

Solved Example 1

Factor `x^3 - 8`

Solution

Rewrite 8 as `2^3`,

`x^3 - 2^3`

Comparing with `a^3 - b^3`, we get

`a = x` and `b = 2`

Applying the formula of difference of cubes: `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get

`(x - 2)(x^2 + x*2 + 2^2)`

Simplifying,

`(x - 2)(x^2 + 2x + 4)`

Solved Example 2

Factor `8a^3 - b^3`

Solution

Rewrite 8 as `2^3`,

`(2a)^3 - b^3`

Comparing with `a^3 - b^3`, we get,

`a = 2a` and `b = b`

Applying the formula `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get

`(2a - b)((2a)^2 + (2a)(b) + b^2)`

Simplifying,

`(2a - b)(4a^2 + 2ab + b^2)`

Solved Example 3

Factor `1 - x^3`

Solution

Rewrite 1 as `1^3` (because `1^3 = 1 * 1 * 1 = 1`).

`1^3 - x^3`

Comparing with `a^3 - b^3`, we get

`a = 1` and `b = x`

Applying the formula `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get

`(1 - x)(1^2 + 1*x + x^2)`

Simplifying,

`(1 - x)(1 + x + x^2)`

Worksheet: Difference of Squares

2016-02-10T10:13:00.002-08:00

Instructions

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

Answer

`x^2 - 16`

Show Answer

`x^4 - 16`

Show Answer

`2x^2 - 50`

Show Answer

`x^3 - xy^2`

Show Answer

`81x^4 - y^4`

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`125x^3 - 36x`

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`162 - 8y^2`

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Solved Examples: Difference of Squares

2016-02-10T10:03:00.000-08:00

Solved Example 1

Factor `x^2 - 4`

Solution

Rewrite 4 as `2^2`

`x^2 – 4 = x^2 – 2^2`

Comparing with `a^2 – b^2`, we get `a = x` and `b = 2`. Applying the formula `a^2 – b^2 = (a + b)(a – b)`, we get

`x^2 – 2^2 = (x + 2)(x – 2)`

Solved Example 2

Factor `x^4 – 81`

Solution

Rewrite `x^4` as `(x^2)^2` and 81 as `9^2`. Thus, we get

`x^4 – 81 = (x^2)^2 – 9^2`

Comparing with `a^2 – b^2`, we get `a = x^2` and `b = 9`. Applying the formula `a^2 – b^2 = (a + b)(a – b)`, we get

`(x^2)^2 – 9^2 = (x^2 + 9)(x^2 – 9)`

Rewrite 9 as `3^2`

`(x^2 + 9)(x^2 – 9) = (x^2 + 9)(x^2 – 3^2)`

Again applying the formula `a^2 – b^2 = (a + b)(a – b)`, we get

`(x^2 + 9)(x^2 – 3^2) = (x^2 + 9)(x + 3)(x – 3)`

Solved Example 3

Factor `2x^2 – 72`

Solution

Factor out 2 from the given expression,

`= 2(x^2 – 36)`

Rewrite 36 as `6^2`.

` = 2(x^2 – 6^2)`

Applying the formula `a^2 – b^2 = (a + b)(a – b)`, we get

`= 2(x + 6)(x – 6)`

Solved Example 4

Factor `a^2 – b^4`

Solution

Rewrite `b^4` as `(b^2)^2` to get

`a^2 – (b^2)^2`

Applying the formula `a^2 – b^2 = (a + b)(a – b)`, we get

`(a + b^2)(a – b^2)`

Solved Example 5

Factor `25x^2 – 36y^2`

Solution

Rewrite 25 as `5^2` and 36 as `6^2`.

`(5x)^2 – (6y)^2`

Applying the formula `a^2 – b^2 = (a + b)(a – b)`, we get

`(5x + 6y)(5x – 6y)`

Solved Example 6

Factor `32x^3 – 50x`

Solution

Factor out `2x` from the given expression,

`= 2x(16x^2 – 25)`

Rewrite 16 as `4^2` and 25 as `5^2`

`= 2x((4x)^2 – 5^2)`

Applying the formula `a^2 – b^2 = (a + b)(a – b)`,

`= 2x(4x + 5)(4x – 5)`

Worksheet: Cube of a Difference

2016-02-10T09:16:00.002-08:00

Instructions

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

Answer

`(4x - y)^3`

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`(5x - 4b)^3`

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`(a/3 + b)^3`

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`(x/2 + 1/y)^3`

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`(3x - 2y)^3`

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`(1 - x)^3`

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`(y - 2)^3`

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`(a/3 + 2/b)^3`

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Solved Examples: Cube of a Difference `(a - b)^3`

2016-02-09T14:36:00.001-08:00

Solved Example 1

Expand `(x -2y)^3`

Solution

Step 1: Compare the given expression with `(a - b)^3`

Comparing `(x - 2y)^3` with `(a - b)^3`, we get

`a = x` and `b = 2y`

Step 2: Apply the formula `(a - b)^3 = a^3 - b^3 - 3ab(a - b)`

Putting `x` in place of `a` and `2y` in place of `b` in the above formula, we get,

`(x - 2y)^3 = x^3 - (2y)^3 - 3(x)(2y)(x - 2y)`

Simplifying the left hand expression,

`(x - 2y)^3 = x^3 - 8y^3 - 6xy(x - 2y)`

`= x^3 - 8y^3 - 6x^2y + 12xy^2`

Solved Example 2

Expand `(3x -2y)^3`

Solution

Step 1: Compare the given expression with `(a - b)^3`

Comparing `(3x - 2y)^3` with `(a - b)^3`, we get

`a = 3x` and `b = 2y`

Step 2: Apply the formula `(a - b)^3 = a^3 - b^3 - 3ab(a - b)`

Putting `3x` in place of `a` and `2y` in place of `b` in the above formula, we get,

`(3x - 2y)^3 = (3x)^3 - (2y)^3 - 3(3x)(2y)(3x - 2y)`

Simplifying the left hand expression,

`(3x - 2y)^3 = 27x^3 - 8y^3 - 18xy(3x - 2y)`

`= 27x^3 - 8y^3 - 54x^2y + 36xy^2`

Solved Example 3

Expand `(x/3 - 2y)^3`

Solution

Step 1: Compare the given expression with `(a - b)^3`

Comparing `(x/3 - 2y)^3` with `(a - b)^3`, we get

`a = x/3` and `b = 2y`

Step 2: Apply the formula `(a - b)^3 = a^3 - b^3 - 3ab(a - b)`

Putting `x/3` in place of `a` and `2y` in place of `b` in the above formula, we get,

`(x/3 - 2y)^3 = (x/3)^3 - (2y)^3 - 3(x/3)(2y)(x/3 - 2y)`

Simplifying the left hand expression,

`(x/3 - 2y)^3 = (x^3)/27 - 8y^3 - 2xy(x - 2y)`

`= (x^3)/27 - 8y^3 - 2x^2y + 4xy^2`

Solved Example 4

Expand `(a/2 - b/3)^3`

Solution

Step 1: Compare the given expression with `(a - b)^3`

Comparing `(a/2 - b/3)^3` with `(a - b)^3`, we get

`a = a/2` and `b = b/3`

Step 2: Apply the formula `(a - b)^3 = a^3 - b^3 - 3ab(a - b)`

Putting `a/2` in place of `a` and `b/3` in place of `b` in the above formula, we get,

`(a/2 - b/3)^3 = (a/2)^3 - (b/3)^3 - 3(a/2)(b/3)(a/2 - b/3)`

Simplifying the left hand expression,

`(a/2 - b/3)^3 = (a^3)/8 - (b^3)/27 - (ab)/2(a/2 - b/3)`

`= (a^3)/8 - (b^3)/27 - (a^2b)/4 + (ab^2)/6`

Solved Example 5

Expand `(x - 1)^3`

Solution

Step 1: Compare the given expression with `(a - b)^3`

Comparing `(x - 1)^3` with `(a - b)^3`, we get

`a = x` and `b = 1`

Step 2: Apply the formula `(a - b)^3 = a^3 - b^3 - 3ab(a - b)`

Putting `x` in place of `a` and `1` in place of `b` in the above formula, we get,

`(x - 1)^3 = x^3 - 1^3 - 3(x)(1)(x - 1)`

Simplifying the left hand expression,

`(x - 1)^3 = x^3 - 1 - 3x(x - 1)`

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

Worksheet: Sum of a Cube `(a + b)^3`

2016-02-09T12:06:00.002-08:00

Instructions

Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers

Question

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`(2x + y)^3`

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`(x + 3y)^3`

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`(3x + 2y)^3`

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`(4a + 7b)^3`

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`(x/2 + 1)^3`

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`(x/3 + y)^3`

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`(x/2 + y/3)^3`

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`(a/3 + b/4)^3`

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`((2x)/3 + (3y)/4)^3`

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`((3a)/2 + (4a)/3)^3`

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Solved Examples: Cube of a Sum `(a + b)^3`

2016-02-09T11:34:00.000-08:00

Solved Example 1

Expand the expression `(x + 2y)^3`

Solution

Compare the given expression with `(a + b)^3`. Thus,

`a = x` and `b = 2y`

Write the formula,

`(a + b)^3 = a^3 + b^3 + 3ab(a + b)`

Put 'x' in place of 'a' and '2y' in place of 'b',

`(x + 2y)^3 = x^3 + (2y)^3 + 3(x)(2y)(x + 2y)`

Simplify the expression,

`(x + 2y)^3 = x^3 + 8y^3 + 6x^2y + 12xy^2`

Solved Example 2

Expand the expression `(3a + 4b)^3`

Solution

Compare the given expression with `(a + b)^3`. Thus,

`a = 3a` and `b = 4b`

Write the formula,

`(a + b)^3 = a^3 + b^3 + 3ab(a + b)`

Put '3a' in place of 'a' and '4b' in place of 'b',

`(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)`

Simplify the expression,

`(3a + 4b)^3 = 27a^3 + 64b^3 + 36ab(3a + 4b)`

`= 27a^3 + 64b^3 + 108a^2b + 144ab^2`

Solved Example 3

Expand the expression `(x + y/2)^3`

Solution

Compare the given expression with `(a + b)^3`. Thus,

`a = x` and `b = y/2`

Write the formula,

`(a + b)^3 = a^3 + b^3 + 3ab(a + b)`

Put `x` in place of 'a' and `y/2` in place of 'b',

`(x + y/2)^3 = (x)^3 + (y/2)^3 + 3(x)(y/2)(x + y/2)`

Simplify the expression,

`(x + y/2)^3 = x^3 + (y^3)/8 + (3xy)/2(x + y/2)`

`= x^3 + (y^3)/8 + (3x^2y)/2 + (3xy^2)/4`

Solved Example 4

Expand the expression `(x/3 + y/4)^3`

Solution

Compare the given expression with `(a + b)^3`. Thus,

`a = x/3` and `b = y/4`

Write the formula,

`(a + b)^3 = a^3 + b^3 + 3ab(a + b)`

Put `x/3` in place of 'a' and `y/4` in place of 'b',

`(x + y/2)^3 = (x)^3 + (y/2)^3 + 3(x)(y/2)(x + y/2)`

Simplify the expression,

`(x + y/2)^3 = x^3 + (y^3)/8 + (3xy)/2(x + y/2)`

`= x^3 + (y^3)/8 + (3x^2y)/2 + (3xy^2)/4`

Solved Examples: Finding the Coordinates of Focus of a Parabola

2016-02-07T15:10:00.002-08:00

Solved Example 1

`y = x^2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 1(x - 0)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 1(x - 0)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 1`

`h = 0`
`k = 0`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(0, 0 + 1/(4*1))`

Simplifying,

`(0, 1/4)`

Thus `(0, 1/4)` is the focus of the parabola `y = x^2`.

Solved Example 2

`y = 4x^2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 4(x - 0)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 4(x - 0)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 4`

`h = 0`
`k = 0`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(0, 0 + 1/(4*4))`

Simplifying,

`(0, 1/(16))`

Thus `(0, 1/(16))` is the focus of the parabola `y = 4x^2`.

Solved Example 3

`y = 5x^2 + 10`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 5(x - 0)^2 + 10`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 5(x - 0)^2 + 10` with `y = a(x - h)^2 + k` we get,

`a = 5`

`h = 0`
`k = 10`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(0, 10 + 1/(4*5))`

Simplifying,

`(0, 201/20)`

Thus `(0, 201/20)` is the focus of the parabola `y = 5x^2 + 10`.

Solved Example 4

`y = 4x^2 + 2x + 2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

Since this equation is not already in the vertex form (unlike the examples above), we'll use the method of completing the square to convert this equation to the vertex form. First, factor out 4 from the LHS,

`y = 4(x^2 + x/2 + 1/2)`

Now, identify the coefficient of 'x', which is `1/2`. Divide it by 2 to get `1/4`. Square it to get `(1/4)^2`. Add and subtract this number from the quadratic expression in the parenthesis,

`y = 4(x^2 + x/2 + (1/4)^2 - (1/4)^2 + 1/2)`

Compare the highlighted part of the equation above with `a^2 + 2ab + b^2` to get `a = x` and `b = 1/4`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`,

`y = 4((x + 1/4)^2 - (1/4)^2 + 1/2)`

Simplify the LHS,

`y = 4((x + 1/4)^2 + 7/16)`

The above equation is now in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 4((x + 1/4)^2 + 7/16)` with `y = a(x - h)^2 + k` we get,

`a = 4`

`h = -1/4`
`k = 7/16`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(-1/4, 7/16 + 1/(4*4))`

Simplifying,

`(-1/4, 1/2)`

Thus `(-1/4, 1/2)` is the focus of the parabola `y = 4x^2 + 2x + 2`.

Solved Example 5

`y = (x - 1)^2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 1(x - 1)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 1(x - 1)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 1`

`h = 1`
`k = 0`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(1, 0 + 1/(4*1))`

Simplifying,

`(1, 1/4)`

Thus `(1, 1/4)` is the focus of the parabola `y = (x - 1)^2`.

Solved Example 6

`y = 2(x + 5)^2 + 6`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 2(x + 5)^2 + 6` with `y = a(x - h)^2 + k` we get,

`a = 2`

`h = -5`
`k = 6`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(-5, 6 + 1/(4*2))`

Simplifying,

`(-5, 49/8)`

Thus `(-5, 49/8)` is the focus of the parabola `y = 2(x + 5)^2 + 6`.

Solved Example 7

`y = (x + 1)(x + 2)`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is in the intercept form. We need to first expand it to the standard form by using FOIL.

`y = x^2+ 3x + 2`

Now we will apply the method of completing the square to convert it to vertex form.

Take the coefficient of `x`, which is 3, divide it by 2 to get `3/2` and then square it to get `(3/2)^2`. Add and subtract `(3/2)^2` from the LHS,

`y = x^2 + 3x + (3/2)^2 - (3/2)^2 + 2`

Compare the highlighted part of the equation above with `a^2 + 2ab + b^2` to get `a = x` and `b = 3/2`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`,

`y = (x + 3/2)^2 - (3/2)^2 + 2`

Simplify the LHS,

`y = (x + 3/2)^2 - 1/4`

The above equation is now in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = (x + 3/2)^2 - 1/4` with `y = a(x - h)^2 + k` we get,

`a = 1`

`h = -3/2`
`k = -1/4`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(-3/2, -1/4 + 1/(4*1))`

Simplifying,

`(-3/2, 0)`

Thus `(-3/2, 0)` is the focus of the parabola `y = (x + 1)(x + 2)`.

Solved Example 8

`x = 4y^2`

Solution

The above equation is different from earlier equations discussed above in that it is not a function. The y-term is squared while the x-term is the dependent variable. It's graph does not pass the vertical line test. Although it is not a function, it can be graphed on a coordinate plane and its focus can be calculated by the formula `(k + 1/(4a), h)`

Step 1: Convert the given equation to vertex form `x = a(y - h)^2 + k`

The given equation is already in the vertex form as it can be written as `x = 4(y - 0)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `x = 4(y - 0)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 4`

`h = 0`
`k = 0`

Step 3: The coordinates of the focus are given by `(k + 1/(4a), h)`

Plug in the values of 'a', 'h' and 'k' to get

`(0 + 1/(4*1), 0)`

Simplifying,

`(1/16, 0)`

Thus `(1/16, 0)` is the focus of the parabola `x = 4y^2`.

Solved Example 9

`x = 5y^2 - 1`

Solution

This equation is same as the above one in that it is not a function but only a quadratic relation.

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `x = 5(y - 0)^2 - 1`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `x = 5(y - 0)^2 - 1` with `y = a(x - h)^2 + k` we get,

`a = 5`

`h = 0`
`k = -1`

Step 3: The coordinates of the focus are given by `(k + 1/(4a), h)`

Plug in the values of 'a', 'h' and 'k' to get

`(-1 + 1/(4*5), 0)`

Simplifying,

`(-19/20, 0)`

Thus `(-19/20, 0)` is the focus of the parabola `x = 5y^2 - 1`.

Solved Example 10

`x = 6y^2 +5y - 1`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

We will apply the method of completing the square to convert it to vertex form. First, factor out 6 from the LHS.

`x =6(y^2 + 5/6y -1/6)`

Take the coefficient of `y`, which is `5/6`, divide it by 2 to get `5/12` and then square it to get `(5/12)^2`. Add and subtract `(5/12)^2` from the LHS,

`x =6( Compare the highlighted part of the equation above with `a^2 + 2ab + b^2` to get `a = y` and `b = 5/12`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`,

`x = 6((y + 5/12)^2 - (5/12)^2 - 1/6)`
Simplify the LHS,

`x = 6(y + 5/12)^2 - 31/6`
The above equation is now in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `x = 6(y + 5/12)^2 - 31/6` with `y = a(x - h)^2 + k` we get,

`a = 6`

`h = -5/12`
`k = -31/6`

Step 3: The coordinates of the focus are given by `(k + 1/(4a), h)`

Plug in the values of 'a', 'h' and 'k' to get

`(-31/6 + 1/(4*6), -5/12)`

Simplifying,

`(-41/8, -5/12)`

Thus `(-41/8, -5/12)` is the focus of the parabola `x = 6y^2 +5y - 1`.

Find Coordinates of Focus of Parabola Method 1

2016-02-07T12:36:00.001-08:00

The following steps can be performed to get the coordinates of the focus of a parabola from its equation:

Convert the equation to vertex form `y = a(x - h)^2 + k`
Identify ‘a’, ‘h’ and ‘k’
Focus of the parabola is `(h, k + 1/(4a))`

The following examples will make it clearer.

Example 1

Find the focus of a parabola with the equation `y = 3(x – 1)^2 + 2`

Solution

Step 1: Convert the equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form. Thus we don’t need to convert it into the vertex form.

Step 2: Identify ‘a’, ‘h’ and ‘k’

Comparing the equation `y = 3(x – 1)^2 + 2` with the vertex form `y = a(x – h)^2 + k` we get

`a = 3`
`h = 1`
`k = 2`

Step3: Focus of the parabola is `(h, k + 1/(4a))`

Plug in the values of ‘a’, ‘h’ and ‘k’ into the above formula to get the coordinates of focus.

`(1, 2 + 1/(4*3))`

Simplify,

`(1, 25/12)`

Thus, the coordinates of the focus of the parabola `y = 3(x – 1)^2 + 2` are `(1, 25/12)`.

Example 2

Find the focus of a parabola with the equation `y = 2x^2 + 3x + 1`

Solution

Step 1: Convert the equation to vertex form `y = a(x - h)^2 + k`

We will do this by using the completing the square method. First, we factor out 2 from the equation.

`y = 2(x^2+ 3/2x + 1/2)`

Now, take the coefficient of ‘x’, which is `3/2`, divide it by 2, whence we get `3/4` and square it, which gives us `(3/4)^2`. Add and subtract `(3/4)^2` from the quadratic expression inside the parenthesis.

`y = 2(x^2 + 3/2x + (3/4)^2 - (3/4)^2 + 1/2)`

Compare the above highlighted part with `a^2 + 2ab + b^2`. Thus we get `a = x` and `b = 3/4`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`.

`y = 2((x + 3/4)^2 - (3/4)^2 + 1/2)`

Simplify,

`y = 2((x + 3/4)^2 - 1/16)`

Remove the parenthesis by multiplying with 2,

`y = 2(x + 3/4)^2 – 1/8`

The above equation is now in vertex form.

Step 2: Identify ‘a’, ‘h’ and ‘k’

Comparing the equation `y = 2(x + 3/4)^2 – 1/8` with the vertex form `y = a(x – h)^2 + k` we get

`a = 2`
`h = -3/4`
`k = -1/8`

Step3: Focus of the parabola is `(h, k + 1/(4a))`

Plug the values of ‘a’, ‘h’ and ‘k’ into the formula for focus to get its coordinates:

`(-3/4, -1/8 + 1/(4*2))`

Simplify,

`(-3/4, 0)`

Thus, the coordinates of the focus of the parabola `y = 2x^2 + 3x + 1` are `(-3/4, 0)`.

Distance Formula Calculator

2016-02-07T12:01:00.001-08:00

Enter the coordinates of two points and click on "Calculate" button to calcualte the distance between them.

Enter Point 1 Coordinates: (, )

Enter Point 2 Coordinates: (, )

Distance = units

Solved Examples: How to find the length of latus rectum of a parabola?

2016-02-05T23:19:00.001-08:00

Example 1: `x^2 = 8y`

on comparing it with `x^2 = 4ay` , we get

`4a = 8`
`a = 2`

and thus length of latus rectum is

Latus Rectum `= 4a = 4 * 2 = 8`

Quite simply, the length of the latus rectum is equal the the coefficient of the y variable here.

But the coefficient can be negative, as in `x^2 = -8y`. The length of the latus rectum remains positive in both cases. That is, it is equal to 8 in both the equations above.

Now suppose you have the equation

Example 2: `y^2 = 12x`

This is comparable with the standard equation

`y^2 = 4ax`

Thus we get

`4a = 12`
`a = 3`

Hence length of latus rectum is

Latus Rectum `= 4 * a = 4 * 3 = 12`

Now suppose we have an equation which is not in the standard form. Let us take, for example

Example 3: `y = 2x^2 + 3x + 1`

Method 1

In order to find the length of the latus rectum for this parabolic equation, we will first convert it into the standard form. (To do that, you need to follow some steps which are discussed in detail here: Change Parabola to Standard Form). It proceeds as follows:

`y = 2(x^2 + 3/2x + 1/2)`

`y = 2(x + 3/2x + (3/4)^2- (3/4)^2 + 1/2)`

The highlighted part above is compared with the formula

`a^2 + 2ab + b^2 = (a + b)^2`

Thus we can write

`y = 2((x + 3/4)^2 - 9/14 + 1/2)`

Simplifying,

`y = 2((x + 3/4)^2 - 1/7)`

Since we notice that the `x` containing term is squared, not `y`, so we know that this equation will be compared to the standard form `x^2 = +/- 4ay`. Hence we will remove all numbers being multiplied to or on the side of the `x` squared term.

Dividing by 2,

`y/2 = (x + 3/4)^2 - 1/7`

Adding 1/7 to both sides,

`y/2 + 1/7 = (x + 3/4)^2`

Method 2

`(x + 3/4)^2 = y/2 + 1/7`

Now we factor out the coefficient of `y`

`(x + 3/4)^2 = 1/2(y + 2/7)`

To make it look more like `x^2 = 4ay`, we will replace `x + 3/4` with `X` and `y + 2/7` with `Y`. Thus we get,

`X^2 = 1/2Y`, where `X = x + 3/4` and `Y = y + 2/7`

This equation of the parabola is same as the original one above, that is, `y = 2x^2 + 3x + 1` and in a different form comparable with the standard form `x^2 = 4ay`.

Now comparing it with `x^2 = 4ay`, we get

`4a = 1/2`
`a = 1/8`

Thus the length of the latus rectum is

Latus Rectum `= 4*a = 4*1/8 = 1/2`

Thus, to find the length of the latus rectum of a parabola you first need to convert the equation into one of its standard forms, and then find the value of `4a` by comparing it with the standard forms.