## Laws of Exponents

Exponents are also called **Powers** or **Indices**

The exponent of a number says **how many times** to use the number in a **multiplication.**

In this example: **8 ^{2} = 8 × 8 = 64**

In words: 8^{2} could be called "8 to the second power", "8 to the power 2" or simply "8 squared"

Try it yourself:

images/exponent-calc.js

So an Exponent saves us writing out lots of multiplies!

### Example: a^{7}

^{7}

a** ^{7}** = a × a × a × a × a × a × a = aaaaaaa

Notice how we wrote the letters together to mean multiply? We will do that a lot here.

### Example: x^{6} = xxxxxx

### The Key to the Laws

Writing all the letters down is the key to understanding the Laws

### Example: x^{2}x^{3} = (xx)(xxx) = xxxxx = x^{5}

Which shows that **x ^{2}x^{3} = x^{5}**, but more on that later!

So, when in doubt, just remember to write down all the letters (as many as the exponent tells you to) and see if you can make sense of it.

### All you need to know ...

The "Laws of Exponents" (also called "Rules of Exponents") come from **three ideas**:

The exponent says how many times to use the number in a multiplication. | |

A negative exponent means divide, because the opposite of multiplying is dividing | |

If you understand those, then you understand exponents!

And all the laws below are based on those ideas.

### Laws of Exponents

Here are the Laws (explanations follow):

Law | Example |
---|---|

x^{1} = x | 6^{1} = 6 |

x^{0} = 1 | 7^{0} = 1 |

x^{-1} = 1/x | 4^{-1} = 1/4 |

x^{m}x^{n} = x^{m+n} | x^{2}x^{3} = x^{2+3} = x^{5} |

x^{m}/x^{n} = x^{m-n} | x^{6}/x^{2} = x^{6-2} = x^{4} |

(x^{m})^{n} = x^{mn} | (x^{2})^{3} = x^{2×3} = x^{6} |

(xy)^{n} = x^{n}y^{n} | (xy)^{3} = x^{3}y^{3} |

(x/y)^{n} = x^{n}/y^{n} | (x/y)^{2} = x^{2} / y^{2} |

x^{-n} = 1/x^{n} | x^{-3} = 1/x^{3} |

And the law about Fractional Exponents: | |

x^{m/n} = n√x^{m}= (n√x ) ^{m} | x^{2/3} = 3√x^{2}= (3√x ) ^{2} |

### Laws Explained

The first three laws above (x^{1} = x, x^{0} = 1 and x^{-1} = 1/x) are just part of the natural sequence of exponents. Have a look at this:

Example: Powers of 5 | |||
---|---|---|---|

.. etc.. | |||

5^{2} | 1 × 5 × 5 | 25 | |

5^{1} | 1 × 5 | 5 | |

5^{0} | 1 | 1 | |

5^{-1} | 1 ÷ 5 | 0.2 | |

5^{-2} | 1 ÷ 5 ÷ 5 | 0.04 | |

.. etc.. |

Look at that table for a while ... notice that positive, zero or negative exponents are really part of the same pattern, i.e. 5 times larger (or 5 times smaller) depending on whether the exponent gets larger (or smaller).

### The law that x^{m}x^{n} = x^{m+n}

With x^{m}x^{n}, how many times do we end up multiplying "x"? *Answer:* first "m" times, then **by another** "n" times, for a total of "m+n" times.

### Example: x^{2}x^{3} = (xx)(xxx) = xxxxx = x^{5}

So, x^{2}x^{3} = x^{(2+3)} = x^{5}

### The law that x^{m}/x^{n} = x^{m-n}

Like the previous example, how many times do we end up multiplying "x"? Answer: "m" times, then **reduce that** by "n" times (because we are dividing), for a total of "m-n" times.

### Example: x^{4}/x^{2} = (xxxx) / (xx) = xx = x^{2}

So, x^{4}/x^{2} = x^{(4-2)} = x^{2}

(Remember that **x**/**x** = 1, so every time you see an **x** "above the line" and one "below the line" you can cancel them out.)

This law can also show you why **x ^{0}=1** :

### Example: x^{2}/x^{2} = **x**^{2-2} = **x**^{0} =1

**x**=

^{2-2}**x**=1

^{0}### The law that (x^{m})^{n} = x^{mn}

First you multiply "m" times. Then you have **to do that "n" times**, for a total of m×n times.

### Example: (x^{3})^{4} = (xxx)^{4} = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x^{12}

So (x^{3})^{4} = x^{3×4} = x^{12}

### The law that (xy)^{n} = x^{n}y^{n}

To show how this one works, just think of re-arranging all the "x"s and "y"s as in this example:

### Example: (xy)^{3} = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x^{3}y^{3}

### The law that (x/y)^{n} = x^{n}/y^{n}

Similar to the previous example, just re-arrange the "x"s and "y"s

### Example: (x/y)^{3} = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x^{3}/y^{3}

### The law that x^{m/n} = n√x^{m} = (n√x )^{m}

OK, this one is a little more complicated!

I suggest you read Fractional Exponents first, so this makes more sense.

Anyway, the important idea is that:

x^{1/n} = The **n-**th Root of x

And so a fractional exponent like 4^{3/2} is really saying to do a **cube** (3) and a **square root** (1/2), in any order.

Just remember from fractions that **m/n = m × (1/n)**:

### Example: x^{(mn)} = x^{(m × 1n)} = (x^{m})^{1/n} = n√x^{m}

The order does not matter, so it also works for **m/n = (1/n) × m**:

### Example: x^{(mn)} = x^{(1n × m)} = (x^{1/n})^{m} = (n√x )^{m}

### Exponents of Exponents ...

What about this example?

4^{32}

We do the exponent at the **top first**, so we calculate it this way:

Start with: | 4^{32} | |

3^{2} = 3×3: | 4^{9} | |

4^{9} = 4×4×4×4×4×4×4×4×4: | 262144 |

### And That Is It!

*If you find it hard to remember all these rules, then remember this:*

you can work them out when you understand the

three ideas near the top of this page:

- The exponent says
**how many times**to use the number in a multiplication - A
**negative exponent**means**divide** - A fractional exponent like
**1/n**means to**take the nth root**: x^{(1n)}= n√x

### Oh, One More Thing ... What if x = 0?

Positive Exponent (n>0) | 0^{n} = 0 | |

Negative Exponent (n<0) | 0^{-n} is undefined (because dividing by 0 is undefined) | |

Exponent = 0 | 0^{0} ... ummm ... see below! |

### The Strange Case of 0^{0}

There are different arguments for the correct value of 0^{0}

0^{0} could be 1, or possibly 0, so some people say it is really "indeterminate":

x^{0} = 1, so ... | 0^{0} = 1 | |

0^{n} = 0, so ... | 0^{0} = 0 | |

When in doubt ... | 0^{0} = "indeterminate" |

323, 2215, 2306, 324, 2216, 2307, 371, 2217, 2308, 2309

ExponentFractional ExponentsPowers of 10Algebra Menu

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Math Expert

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(A) \(x^3 − y^3\)

(B) \((x^2 + y^2)^3\)

(C) \((x^3 + y^3)^3\)

(D) \((x^3 − y^3)^2\)

(E) \((x^3 + y^3)^2\)

Source: Nova GMAT

Difficulty Level: 550

Originally posted by Bunuel on Jun 01, 2017 7:00 pm.

Last edited by Sajjad1994 on Jul 08, 2019 12:35 pm, edited 1 time in total.

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Bunuel wrote:

\(4(xy)^3 + (x^3 − y^3)^2 =\)

(A) \(x^3 − y^3\)

(B) \((x^2 + y^2)^3\)

(C) \((x^3 + y^3)^3\)

(D) \((x^3 − y^3)^2\)

(E) \((x^3 + y^3)^2\)

\(4(xy)^3 + (x^3 − y^3)^2 =\) - expand the Square of a Difference:

\(4(xy)^3\) + \(x^6 - (2)(xy)^3 + y^6\) =

\(x^6 + (2)(xy)^3 + y^6\) = the expanded version of Square of a Sum ==>

\((x^3 + y^3)^2\)

Answer

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We have

\(4(xy)^3 + (x^3 - y^3)^2 = > 4(xy)^3 + x^6 - 2(xy)^3 + y^6 = x^6 + 2(xy)^3 + y^6 = (x^3+y^3)^2\)

Therefore, the answer is option E

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This can also be solved by plugging in values. Lets assume x = y = 1.

So 4(xy)^3 + (x^3 − y^3)^2 = 4 *(1*1)^2 + (1-1)^2 = 4.

Thus we have to look for an option where when we put x=1 and y=1, we should get our answer as '4'.

Check A) 1-1 = 0

B) (1+1)^3 = 8

C) (1+1)^3 = 8

D) (1-1)^2 = 0

E) (1+1)^2 = 4

Only E satisfies. Thus E answer

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amanvermagmat wrote:

This can also be solved by plugging in values. Lets assume x = y = 1.

So 4(xy)^3 + (x^3 − y^3)^2 = 4 *(1*1)^2 + (1-1)^2 = 4.

Thus we have to look for an option where when we put x=1 and y=1, we should get our answer as '4'.

Check A) 1-1 = 0

B) (1+1)^3 = 8

C) (1+1)^3 = 8

D) (1-1)^2 = 0

E) (1+1)^2 = 4

Only E satisfies. Thus E answer

4x^3*y^3 + x^6 - 2(x^3- y^3) + y^6. On opening up the brackets we get, 4x^3*y^3 + x^6 - 2x^3y^3 + y^6. Cancel out 4x^3y^3 and 2x^3y^3. Finally we get, x^6 + 2x^3y^3 +y^6. Therefore, the answer is E.

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WE:**Engineering (Consumer Products)**

= 4(xy)^3 + x^6 + y^6 -2(xy)^3

= x^6 + y^6 +2(xy)^3

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

Option E is should be the correct answer, waiting for OA

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=4(xy)^3 + (x^6 - 2(xy)^3 + y^6)

=2(xy)^3 +x^6 +y^6

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

Hence, the answer should be E.

I would appreciate a kudos if you liked my solution!

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Let x=y=1

The expression \(4(xy)^3 + (x^3 − y^3)^2\) gives us a value of 4

Evaluating the answer choices, Only \((x^3+y^3)^2\)(Option E) gives us the same value

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4(xy)^3+x^6-2x^3y^3+y^3

=x^6+y^6+2x^3y^3

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

Hence Option E

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Bunuel wrote:

\(4(xy)^3 + (x^3 − y^3)^2 =\)

(A) \(x^3 − y^3\)

(B) \((x^2 + y^2)^3\)

(C) \((x^3 + y^3)^3\)

(D) \((x^3 − y^3)^2\)

(E) \((x^3 + y^3)^2\)

4(xy)^3 + (x^3 - y^3)^2

4x^3y^3 + x^6 - 2x^3y^3 + y^6

x^6 + 2x^3y^3 + y^6

(x^3 + y^3)^2

Answer: E

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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink]

Sep 21, 2018 7:54 pmSours: https://gmatclub.com/forum/4-xy-3-x-3-y-241754.html

## Algebra Calculator Tutorial

This is a tutorial on how to use the **Algebra Calculator**, a step-by-step calculator for algebra.

### Solving Equations

First go to the Algebra Calculator main page. In the Calculator's text box, you can enter a math problem that you want to calculate.

For example, try entering the equation 3x+2=14 into the text box.

After you enter the expression, Algebra Calculator will print a step-by-step explanation of how to solve 3x+2=14.

### Examples

To see more examples of problems that Algebra Calculator understands, visit the Examplespage. Feel free to try them now.### Math Symbols

If you would like to create your own math expressions, here are some symbols that Algebra Calculator understands:

**+** (Addition) **-** (Subtraction) ***** (Multiplication) **/** (Division) **^** (Exponent: "raised to the power")

### Graphing

To graph an equation, enter an equation that starts with "y=" or "x=". Here are some examples: y=2x^2+1, y=3x-1, x=5, x=y^2.

To graph a point, enter an ordered pair with the x-coordinate and y-coordinate separated by a comma, e.g., (3,4).

To graph two objects, simply place a semicolon between the two commands, e.g., y=2x^2+1; y=3x-1.

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Algebra Calculator can simplify polynomials, but it only supports polynomials containing the variable x.

Here are some examples: x^2 + x + 2 + (2x^2 - 2x), (x+3)^2.

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To evaluate an expression containing x, enter the expression you want to evaluate, followed by the @ sign and the value you want to plug in for x. For example the command 2x @ 3 evaluates the expression 2x for x=3, which is equal to 2*3 or 6.

Algebra Calculator can also evaluate expressions that contain variables x and y. To evaluate an expression containing x and y, enter the expression you want to evaluate, followed by the @ sign and an ordered pair containing your x-value and y-value. Here is an example evaluating the expression xy at the point (3,4): xy @ (3,4).

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As an example, suppose we solved 2x+3=7 and got x=2. If we want to plug 2 back into the original equation to check our work, we can do so: 2x+3=7 @ 2. Since the answer is right, Algebra Calculator shows a green equals sign.

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To check an answer to a system of equations containing x and y, enter the two equations separated by a semicolon, followed by the @ sign and an ordered pair containing your x-value and y-value. Example: x+y=7; x+2y=11 @ (3,4).

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