Properties of Exponents

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Taking advantage of all the properties of powers or laws of exponents

From time to time, we will come across exercises in which we must use all the properties of powers together.
As soon as you have the exercise, try to first get rid of the parentheses according to the properties of powers and then, apply these properties to the corresponding terms, one after the other.

All the properties of powers or laws of exponents are:
amΓ—an=a(m+n)a^m\times a^n=a^{(m+n)}
aman=a(mβˆ’n)\frac {a^m}{a^n} =a^{(m-n)}
(aΓ—b)n=anΓ—bn(a\times b)^n=a^n\times b^n
(ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}
(an)m=a(nβˆ—m)(a^n )^m=a^{(n*m)}
a0=1a^0=1
When a≠0a≠0
aβˆ’n=1ana^{-n}=\frac {1}{a^n}

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\( (3\times4\times5)^4= \)

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Properties of Exponentiation Examples

Example of an exercise for which we must use all the properties of powers or laws of exponents:

Example 1

(3x)βˆ’3Γ—(3βˆ’3)4Γ—xβˆ’2xβˆ’4(\frac{3}{x})^{-3}\times \frac{(3^{-3})^4\times x^{-2}}{x^{-4}}

We know that this exercise might scare you a lot but, trust us, it includes everything you have already studied.
The way to solve it is simply to try to figure out what you can do with the power properties you have learned.
Shall we?
Observing the exercise, we will see that there are terms that are in parentheses. According to the order of mathematical operations, parentheses are in the first place, so we will start with them.
Let's look at the first term with parentheses: (3x)βˆ’3(\frac{3}{x})^{-3}
According to this property:Β (ab)n=anbn(\frac {a}{b})^n=\frac {a^n}{b^n}

It will give us:

(3x)βˆ’3=3βˆ’3xβˆ’3(\frac{3}{x})^{-3}=\frac{3^{-3}}{x^{-3}}
Let's move on to the second term with parentheses: (3βˆ’3)4(3^{-3})^4

According to this property:
(an)m=a(nΓ—m) (a^n)^m=a^{(n\times m)}

It will give us:Β 
(3βˆ’3)4=3βˆ’3=4=3βˆ’12(3^{-3})^4=3^{-3=4}=3^{-12}

Great! We have gotten rid of the parentheses! Now we will write the exercise according to what we have reached so far:

3βˆ’3xβˆ’3Γ—3βˆ’12Γ—xβˆ’2xβˆ’4= \frac{3^{-3}}{x^{-3}}\times\frac{3^{-12}\times x^{-2}}{x^{-4}}=

Pay attention that, between our two terms there is a multiplication operation.
Therefore, we can join them into a single term by writing them exactly as they appear, but under a fraction bar.
Let's not forget the multiplication between them.
We will obtain:
3βˆ’3Γ—3βˆ’12Γ—xβˆ’2xβˆ’3Γ—xβˆ’4\frac{3^{-3}\times 3^{-12}\times x^{-2}}{x^{-3}\times x^{-4}}
We will realize that, in the exercise, there are hidden equal bases with multiplication operation between them. Consequently, we can add the concerning powers.
According to the following property:Β 
amΓ—an=a(m+n) a^m\times a^n=a^{(m+n)}

We will obtain:
3βˆ’15Γ—xβˆ’2xβˆ’7 \frac{3^{-15}\times x^{-2}}{x^{-7}}

It's looking much better, right?
Let's continue.
We will realize that, in the exercise we have now, there is a fraction with equality of bases. Thus, we can subtract the concerning powers.
According to the following property:
aman=a(mβˆ’n)\frac {a^m}{a^n} =a^{(m-n)}

We will obtain:
3βˆ’15Γ—x5= 3^{-15}\times x^5=

Pay attention, we subtract the powers βˆ’2-2 and βˆ’5-5:
When you subtract a negative from a negative, the result is positive.

We have reached the solution, but note that, you might be asked to show the result only with positive powers.
We can transform the negative power βˆ’15-15 to positive according to the following property:
aβˆ’n=1ana^{-n}=\frac {1}{a^n}

We will obtain:
1315Γ—x5 \frac{1}{3^{15}}\times x^5

We will multiply and obtain:
x5315\frac{x^5}{3^{15}}


The more you practice using these properties or laws separately, the easier it will be for you to apply them together and know which one to use first in a complex exercise.

Here is an example of a complex exercise that requires the use of several properties together.

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Example 2

(5XX+2Y)βˆ’3Γ—(Xβˆ’2)βˆ’4Γ—Yβˆ’1Γ—50Y2= (\frac{5X^{X+2}}{Y})^{-3}\times\frac{(X^{-2})^{-4}{\times Y^{-1}}\times5^0}{Y^2}=

Wow... You might be looking at the exercise and wondering if you have any idea where to start... That's totally natural since it really looks threatening and includes many negative signs, fractions, and parentheses.

But fear not, we are here to solve it together and we will give you recommendations so you can succeed with similar exercises that you will receive from now on, even in exams.

Our first recommendation for this type of exercises is not to torment yourself with fractions and negatives, look at the terms that are in parentheses.

In the order of mathematical operations, this is the first of all, and it is important that we get rid of the parentheses.

Let's start with the first expression and remember the property of the power of a quotient that says that each term is raised to the power separately.

We will realize that in the numerator we already have a power over the base XX so we will use the property of the power of a power.

Let's not forget to apply the power also over the coefficient of the XX.

We will do it and obtain:

5βˆ’3Γ—X(X+2)Γ—(βˆ’3)Yβˆ’3Γ—(Xβˆ’2)βˆ’4Γ—Yβˆ’1Γ—50Y2= \frac{5^{-3}\times X^{(X+2)\times(-3)}}{Y^{-3}}\times\frac{(X^{-2})^{-4}\times Y^{-1}\times5^0}{Y^2}=

We will continue opening the parentheses in the exponent of the XX in the numerator of the first term and obtain:

5βˆ’3Γ—Xβˆ’3Xβˆ’6Yβˆ’3Γ—(Xβˆ’2)βˆ’4Γ—Yβˆ’1Γ—50Y2= \frac{5^{-3}\times X^{-3X-6}}{Y^{-3}}\times \frac{(X^{-2})^{-4}\times Y^{-1}\times 5^0}{Y^2}=

We know you still can't see how all this will be solved, but hey, be patient. We have only taken the first step.

Now we will move on to the second term and there we will also remove the parentheses. We will apply the property of power of a power on the XX and obtain:

5βˆ’3Γ—Xβˆ’3Xβˆ’6Yβˆ’3Γ—X8Γ—Yβˆ’1Γ—50Y2= \frac{5^{-3}\times X^{-3X-6}}{Y^{-3}}\times \frac{X^{8}\times Y^{-1}\times 5^0}{Y^2}=

Now we will realize that we have a certain term raised to 00, and we already know that, any term raised to 00 is equal to 11. Consequently, we can ignore it and write the exercise this way:

5βˆ’3Γ—Xβˆ’3Xβˆ’6Yβˆ’3Γ—X8Γ—Yβˆ’1Y2= \frac{5^{-3}\times X^{-3X-6}}{Y^{-3}}\times \frac{X^{8}\times Y^{-1}}{Y^2}=

It's looking a bit more organized now, isn't it?

Let's continue.

Our second recommendation is to combine the fractions, especially if there is multiplication between them. Generally, after removing the parentheses, this will be the next step.

Indeed, now we want to end up with a single fraction. We will notice that there is multiplication between them, so we can do it very easily.

Notice that, when we multiply like bases we can add the exponents and obtain a single base with a single exponent.

We will calculate and obtain:

5βˆ’3Γ—Xβˆ’3X+2Γ—Yβˆ’1Yβˆ’1= \frac{5^{-3}\times X^{-3X+2}\times Y^{-1}}{Y^{-1}}=

Ah! This is already starting to look like something familiar where we can act without fear.

We will realize that, both in the numerator and in the denominator, we have an identical base and exponent that we can reduce. Obviously, we will do it and, in this way, dissolve the fraction.

We will receive:

5βˆ’3Γ—Xβˆ’3X+2= 5^{-3}\times X^{-3X+2}=

Excellent! Now we will use the property of the negative exponent, we will convert the terms into a fraction according to the rules. Remember that if the negative exponent is an expression we must alter the signs of each of the terms of the expression.

We will receive:

153Γ—1X3Xβˆ’2= \frac{1}{5^3}\times \frac{1}{X^{3X-2}}=

We can combine the fractions and arrive at:

1125X3Xβˆ’2 \frac{1}{125X^{3X-2}}

We're done!

See? Quite simple, isn't it?

The trick is to use the properties you learned and understand which one you can use in each case.

Rest assured that, if you work slowly and carefully, you will reach the correct result.


If you are interested in this article, you might also be interested in the following articles:

  • Powers
  • The Rules of Exponentiation
  • Division of Powers with the Same Base
  • Power of a Multiplication
  • Power of a Quotient
  • Power of a Power
  • Power with Zero Exponent
  • Powers with a Negative Integer Exponent
  • Exponentiation of Integers

In the Tutorela blog, you will find a variety of articles about mathematics.


Properties of Powers Exercises

Exercise 1

Solve the following exercise:

23Γ—24+(43)2+2523= 2^3\times2^4+(4^3)^2+\frac{2^5}{2^3}=

Solution:

We start by solving the product: 23Γ—24 2^3\times2^4

According to the multiplication of powers with the same base, we will add the power coefficients, 3+4 3+4

Then the expression (43)2 \left(4^3\right)^2 ,

According to the power of a power property, we can multiply the powers 2Γ—3=6 2\times3=6 .

Finally, in the expression 2523 \frac{2^5}{2^3} we use the property of the quotient of power with the same base, and through this, we subtract the coefficients, 5βˆ’3=2 5-3=2

Therefore

Answer:

22+27+46 2^2+2^7+4^6

NOTE: In this exercise, we can write 4 4 as 22 2^2 and obtain: 22+27+2122^2 +2^7 +2^{12}


Do you know what the answer is?

Exercise 2

Solve the following exercise:

((9xyz)3)4+(ay)x= \left(\left(9xyz\right)^3\right)^4+\left(a^y\right)^x=

Solution:

In this exercise, there are two central laws, the power of a power property and the power of a product.

Since both are essentially multiplication operations, it is possible to use the commutative law in this case and start by multiplying the coefficients.

3Γ—4=12 3\times4=12

We use the first law also for the second factor of the exercise.

(xΓ—y) (x\times y)

And then we will use the second law to give each of the numbers in parentheses the power, and so it results:

912Γ—x12Γ—y12Γ—z12+ayx 9^{12}\times x^{12}\times y^{12}\times z^{12}+a^{yx}

Answer:

912Γ—x12Γ—y12Γ—z12+ayx 9^{12}\times x^{12}\times y^{12}\times z^{12}+a^{yx}


Exercise 3

Solve the following exercise:

X3Γ—X2:X5+X4 X^3\times X^2:X^5+X^4

Solution:

  • First, we solve the multiplications and divisions from left to right.
  • Then addition and subtraction
  • Also, we use the power laws (Power of a product)

X3Γ—X2:X5+X4= X^3\times X^2:X^5+X^4=

We add the powers that have multiplication with the same base.

X2+3=X5 X^{2+3}=X^5

X5:X5+X4= X^5:X^5+X^4=

X5:X5=1 X^5:X^5=1

1+X4 1+X^4

Answer:

1+X4 1+X^4


Check your understanding

Exercise 4

Solve the following exercise:

(4Γ—9Γ—11)a (4\times9\times11)^a

Solution:

According to the power property, when we encounter an expression in which the power value appears throughout the product or in the entire exercise where there are only multiplication operations among the members (using parentheses throughout the expression), we can take the power value and apply it to each product

That is, each of the products is raised to the power.

Therefore 4a9a11a 4^a9^a11^a

Answer:

4a9a11a 4^a9^a11^a


Exercise 5

Solve the following exercise:

(x2Γ—3)2= (x^2\times3)^2=

Solution:

In this task, there is the use of two laws, both the multiplication of powers and the power of a power. Each of the products inside the parentheses receives the external power, since they have different bases and a multiplication operation between them. The power inside the parentheses is multiplied by the power outside of it, according to the law of a power of a power.

Therefore:

(x2)2Γ—32=x4Γ—9=9x4 (x^2)^2 \times 3^2=x^4 \times 9 = 9x^4

Answer:

(x2Γ—3)2=9x4 (x^2\times3)^2=9x^4


Do you think you will be able to solve it?

Review Questions

What are the properties of powers?

The properties of exponents are the laws that tell us how to work with exponents. Depending on the form of the expression or the exercise, we may be able to use some of these laws.


How are the properties of powers applied?

Depending on the exercise, we can apply a property of the exponent. There are properties for multiplication and division of powers with the same base, properties for power of a power, power of products and quotients, as well as the law for zero exponent or negative integer exponent.


What are the power properties and examples?

The properties of exponents are the following:

  • Product of powers with the same base: amβ‹…an=am+n a^m\cdot a^n=a^{m+n}

Example: 32β‹…35=32+5=37 3^2\cdot3^5=3^{2+5}=3^7

  • Quotient of powers with the same base: aman=amβˆ’n \frac{a^m}{a^n}=a^{m-n}

Example: 5854=58βˆ’4=54 \frac{5^8}{5^4}=5^{8-4}=5^4

  • Power of products: (aβ‹…b)n=anbn \left(a\cdot b\right)^n=a^nb^n

Example: (3β‹…5)2=32β‹…52 \left(3\cdot5\right)^2=3^2\cdot5^2

  • Power of quotients: (ab)n=anbn \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}

Example: (95)3=9353 \left(\frac{9}{5}\right)^3=\frac{9^3}{5^3}

  • Power of a power: (am)n=amβ‹…n \left(a^m\right)^n=a^{m\cdot n}

Example: (62)5=62β‹…5=610 \left(6^2\right)^5=6^{2\cdot5}=6^{10}

  • Power with zero exponent: a0=1 a^0=1

Example: 30=1 3^0=1

  • Power with negative integer exponent: aβˆ’n=1an a^{-n}=\frac{1}{a^n}

Example:2βˆ’3=123 2^{-3}=\frac{1}{2^3}


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