Xy 3 4

Xy 3 4 DEFAULT

Laws of Exponents

Exponents are also called Powers or Indices

8 to the Power 2

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

In this example: 82 = 8 × 8 = 64

In words: 82 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: a7

a7 = 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: x6 = xxxxxx

The Key to the Laws

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

Example: x2x3 = (xx)(xxx) = xxxxx = x5

Which shows that x2x3 = x5, 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:

pencil paperThe exponent says how many times to use the number in a multiplication.
  
turn overA negative exponent means divide, because the opposite of multiplying is dividing
  
pie slice

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):

LawExample
x1 = x61 = 6
x0 = 170 = 1
x-1 = 1/x4-1 = 1/4


xmxn = xm+nx2x3 = x2+3 = x5
xm/xn = xm-nx6/x2 = x6-2 = x4
(xm)n = xmn(x2)3 = x2×3 = x6
(xy)n = xnyn(xy)3 = x3y3
(x/y)n = xn/yn(x/y)2 = x2 / y2
x-n = 1/xnx-3 = 1/x3
And the law about Fractional Exponents:
xm/n  = n√xm
           = (n√x )m
x2/3  = 3√x2
           = (3√x )2

Laws Explained

The first three laws above (x1 = x, x0 = 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.. exponent 5x larger smaller
521 × 5 × 525
511 × 55
5011
5-11 ÷ 50.2
5-21 ÷ 5 ÷ 50.04
 .. 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 xmxn = xm+n

With xmxn, 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: x2x3 = (xx)(xxx) = xxxxx = x5

So, x2x3 = x(2+3) = x5

The law that xm/xn = xm-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: x4/x2 = (xxxx) / (xx) = xx = x2

So, x4/x2 = x(4-2) = x2

(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 x0=1 :

Example: x2/x2 = x2-2 = x0 =1

The law that (xm)n = xmn

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

Example: (x3)4 = (xxx)4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x12

So (x3)4 = x3×4 = x12

The law that (xy)n = xnyn

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) = x3y3

The law that (x/y)n = xn/yn

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) = x3/y3

The law that   xm/n  = n√xm  = (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:

x1/n = The n-th Root of x

And so a fractional exponent like 43/2 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)  =  (xm)1/n  =  n√xm

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

Example: x(mn)  =  x(1n × m)  =  (x1/n)m  =  (n√x )m

Exponents of Exponents ...

What about this example?

432

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

Start with: 432
32 = 3×3: 49
49 = 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) 0n = 0
Negative Exponent (n<0) 0-n is undefined (because dividing by 0 is undefined)
Exponent = 0 00 ... ummm ... see below!

 

The Strange Case of 00

There are different arguments for the correct value of 00

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

question markx0 = 1, so ... 00 = 1
0n = 0, so ... 00 = 0
When in doubt ...00 = "indeterminate"

 

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

ExponentFractional ExponentsPowers of 10Algebra Menu

Sours: https://www.mathsisfun.com/algebra/exponent-laws.html

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error LogLearn more

Hello Guest!

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join 700,000+ members and get the full benefits of GMAT Club

Registration gives you:

  • Tests

    Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan Prep. All are free for GMAT Club members.

  • Applicant Stats

    View detailed applicant stats such as GPA, GMAT score, work experience, location, application status, and more

  • Books/Downloads

    Download thousands of study notes, question collections, GMAT Club’s Grammar and Math books. All are free!

and many more benefits!

Math Expert

Joined: 03 Sep 2009

Posts: 80296

Own Kudos [?]: 434072 [4]

Given Kudos: 55874

Premium MemberCAT TestsForum Quiz ModeTop Member of the Month
4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Updated on: Jul 08, 2019 12:35 pm

00:00

Difficulty:

15% (low)

Question Stats:

84% (01:29) correct 16%(02:00)wrong based on 97 sessions

HideShow timer Statistics

\(4(xy)^3 + (x^3 − y^3)^2 =\)


(A) \(x^3 − y^3\)

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

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

(D) \((x^3 − y^3)^2\)

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

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.

Added Source

Senior SC Moderator

Joined: 23 May 2016

Posts: 5046

Own Kudos [?]: 23880 [0]

Given Kudos: 8450

Premium MemberCAT TestsForum Quiz ModeTop Member of the Month
Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 01, 2017 7:52 pm

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

Current Student

Joined: 04 Jan 2016

Posts: 162

Own Kudos [?]: 101 [1]

Given Kudos: 422

Location: United States (NY)

Schools: W. P. Carey Arizona State '18, Goizueta '18, Carlson '18, Wharton '20 (A)

GMAT 1:620 Q44 V32

GMAT 2:600 Q48 V25

GMAT 3:660 Q42 V39

GPA: 3.48

GMAT ToolKit User
Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 04, 2017 7:44 pm

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

Hit the K

Retired Moderator

Joined: 22 Aug 2013

Posts: 1317

Own Kudos [?]: 1981 [1]

Given Kudos: 459

Location: India

Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 05, 2017 7:42 am

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

avatar

Intern

Intern

Joined: 23 May 2017

Posts: 5

Own Kudos [?]: 0 [0]

Given Kudos: 0

Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 05, 2017 10:18 am

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.

Manager

Manager

Joined: 10 Dec 2015

Posts: 103

Own Kudos [?]: 47 [0]

Given Kudos: 48

Location: India

Concentration: General Management, Operations

Schools: IIMC (A)

GMAT 1:700 Q49 V36

GPA: 3.5

WE:Engineering (Consumer Products)

Reviews Badge
Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 07, 2017 11:40 am
4(xy)^3 + (x^3-y^3)^2
= 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 :)

Intern

Intern

Joined: 19 Jun 2015

Posts: 48

Own Kudos [?]: 48 [0]

Given Kudos: 24

GMAT 1:760 Q49 V44

Reviews Badge
Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 07, 2017 12:21 pm

=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!

Senior PS Moderator

Joined: 26 Feb 2016

Posts: 3169

Own Kudos [?]: 4189 [0]

Given Kudos: 46

Location: India

GPA: 3.12

Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Jun 07, 2017 12:27 pm

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

Manager

Manager

Joined: 19 Jul 2017

Posts: 79

Own Kudos [?]: 22 [0]

Given Kudos: 85

Location: India

Concentration: General Management, Strategy

GPA: 3.5

Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Aug 28, 2017 7:21 am

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

Target Test Prep Representative

Joined: 04 Mar 2011

Status:Head GMAT Instructor

Affiliations: Target Test Prep

Posts: 2807

Own Kudos [?]: 4046 [0]

Given Kudos: 1518

Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Aug 31, 2017 5:21 pm

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
User avatar

Non-Human User

Joined: 09 Sep 2013

Posts: 20747

Own Kudos [?]: 617 [0]

Given Kudos: 0

Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Sep 21, 2018 7:54 pm

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

GMAT Club Bot

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

Sep 21, 2018 7:54 pm

Sours: https://gmatclub.com/forum/4-xy-3-x-3-y-241754.html
  1. Functional ambulation profile
  2. Legionella pronunciation
  3. Rat coloring sheets
  4. Flat illustration portrait

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.

Polynomials

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.


Evaluating Expressions

Algebra Calculator can evaluate expressions that contain the variable x.

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).

Checking Answers For Solving Equations

Just as Algebra Calculator can be used to evaluate expressions, Algebra Calculator can also be used to check answers for solving equations containing x.

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.

If we instead try a value that doesn't work, say x=3 (try 2x+3=7 @ 3), Algebra Calculator shows a red "not equals" sign instead.

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).


Tablet Mode

If you are using a tablet such as the iPad, enter Tablet Mode to display a touch keypad.


Related Articles



Sours: https://www.mathpapa.com/calc/tutorial/
Complete the table to draw graph of the equations x + y = 3 and x - y = 4
Tick mark Image
\left. \begin{array} { l } { x - y = 3 } \\ { 4 x + y = 17 } \end{array} \right.
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
Add y to both sides of the equation.
Substitute y+3 for x in the other equation, 4x+y=17.
Subtract 12 from both sides of the equation.
Substitute 1 for y in x=y+3. Because the resulting equation contains only one variable, you can solve for x directly.
The system is now solved.

Similar Problems from Web Search

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
Add y to both sides of the equation.
Sours: https://mathsolver.microsoft.com/en/solve-problem/%60left.%20%60begin%7Barray%7D%20%7B%20l%20%7D%20%7B%20x%20-%20y%20%3D%203%20%7D%20%60%60%20%7B%204%20x%20%2B%20y%20%3D%2017%20%7D%20%60end%7Barray%7D%20%60right.

3 4 xy

.

xy 3 SE install video--how to install xy-3 se?

.

You will also like:

.



1149 1150 1151 1152 1153