# Xy 3 4

## 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: 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.

### 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:

 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):

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..
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 :

### 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:

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

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

### 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:

### Exponents of Exponents ...

432

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

### 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

• 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":

 x0 = 1, so ... 00 = 1 0n = 0, so ... 00 = 0 When in doubt ... 00 = "indeterminate"

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Sep 21, 2018 7:54 pm

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## 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.

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

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

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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).

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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
\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.

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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.

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