Negative Exponents

πŸ†Practice powers with negative integer exponent

When we see any number(positive or negative) raised to a negative power we can convert the expression into a fraction and we will do it as follows:
the numerator will be 11, the denominator will be the base of the power as seen in the original exercise, but now, with a positive exponent.
That is to say, in the denominator we will invert the exponent to positive.
Pay attention, we will not modify the sign of the base of the potentiation even if it is negative.
Property formula:
aβˆ’n=1ana^{-n}=\frac {1}{a^n}
This property also applies to algebraic expressions.

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Test yourself on powers with negative integer exponent!

einstein

\( (\frac{1}{4})^{-1} \)

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We will see several examples of negative exponents

Example 1

4βˆ’3=4^{-3}=
We can see that the exponent is a negative number. Therefore, we will convert the expression to a fraction in this way: the numerator will be and in the denominator we will place the base of the power with positive exponent.

11
That is to say:

143\frac{1}{4^3}


If we have a fraction with a 11 in the numerator and a base with some expression in the denominator, we must add a subtraction sign to the exponent in the denominator to convert it into a base with a negative exponent. Only in this way we will obtain the correct exponent.


Example 2

142βˆ’X= \frac{1}{4^{2-X}}=

In order to write the expression as a power with negative exponent we must add a negative sign outside the exponent in the denominator as follows:

4βˆ’(2βˆ’X)= 4^{-(2-X)}=

Then we will take out the parentheses and we will obtain:

4βˆ’2+X= 4^{-2+X}=

Another way to do this is to modify the sign of each term.


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Example 3

3β‹…5Xβˆ’7= 3\cdot5X^{-7}=

In the expression we have a negative exponent. We will begin by separating the XX from its coefficient 55.

We have:

3β‹…5β‹…Xβˆ’7= 3\cdot5\cdot X^{-7}=

Now it is clearer to us that to proceed we will have to convert the XX into fraction in the way we learned. We will do it and we will obtain:

3β‹…5β‹…1X7= 3\cdot5\cdot\frac{1}{X^7}=

Great. We will multiply the terms and we will get:

15X7 \frac{15}{X^7}


Example 4

2βˆ’44βˆ’5 \frac{2^{-4}}{4^{-5}}

I'm sure you're saying, "Hey, how am I going to solve this exercise?

That's exactly what we're here for. We will do it slowly and safely.

Remember that the properties do not change: when there is a base with negative exponent it becomes a fraction just as we have learned. We will convert each term to a fraction and get:

124145= \frac{\frac{1}{2^4}}{\frac{1}{4^5}}=

Then, simply use the property of division of fractions:

124:145 \frac{1}{2^4}:\frac{1}{4^5}

We will convert it to a multiplication operation and invert the fraction by which we divide. We will obtain:

124β‹…451= \frac{1}{2^4}\cdot\frac{4^5}{1}=

We will solve and get:

4524= \frac{4^5}{2^4}=

We can express the 44 as 22 2^2 , then we apply the quotient law of exponents with the same base and we will have:

21024= \frac{2^{10}}{2^4}=

Since we have the same base we can subtract the exponents and we will get:

26=64 2^6=64

Attention!

We could have solved the exercise in a much simpler way if we had thought before to convert the 4 to 22 2^2 .

If so, we would have created from the beginning a fraction with equal bases and we would have subtracted the exponents.

Shall we try?

Let us recall the original exercise:

2βˆ’44βˆ’5= \frac{2^{-4}}{4^{-5}}=

Now let's put the 4 as 22 2^2 We will have:

2βˆ’4(22)βˆ’5= \frac{2^{-4}}{(2^2)^{-5}}=

Now let's apply to the denominator the power property of a power and arrive at:

2βˆ’4(2)βˆ’10= \frac{2^{-4}}{(2)^{-10}}=

Good. Now we can subtract the exponents since we have equality of bases.

Let's remember that when we multiply negative numbers the result is positive.

We will get:

26=64 2^6=64

And, as you see, we have obtained the same result by a shorter path.


What happens when there is a fraction that is raised totally to a negative exponent?

It is very simple, we will invert the numerator with the denominator and transform the exponent to positive.

Do you know what the answer is?

Let's see an example

(24)βˆ’3= (\frac{2}{4})^{-3}=

We see that we have a negative exponent that applies to the whole fraction.

Therefore, we will invert the numerator with the denominator, transform the exponent to positive and obtain:

(42)3= (\frac{4}{2})^3=

Tip: before rushing to apply the exponent to each term it is convenient to look at what expression is between the parentheses. Clearly we have a 22.

Therefore, it will give us:

23=8 2^3=8


Exercises for negative exponents

Exercise 1

Solve the following exercise:

(14)βˆ’1 (\frac{1}{4})^{-1}

Solution:

When we have a negative power, we convert the numerator and denominator, thus 14 \frac{1}{4} converting41 \frac{4}{1} , i.e., the answer is.4 4

Answer:

4 4


Check your understanding

Exercise 2

Solve the following exercise:

5βˆ’2 5^{-2}

Solution:

According to the property, when we have a whole number with a power that is negative, the number will become a fraction of the original number, when the power will affect the denominator (the one that was the original number).

In other words 5βˆ’2=152 5^{-2}=\frac{1}{5^2}

Now all that remains is to solve for the power in the denominator.

Answer:

125 \frac{1}{25}


Exercise 3

Solve the following exercise:

[(17)βˆ’1]4= [(\frac{1}{7})^{-1}]^4=

Solution:

In this exercise there are two parts, and we start by solving according to the order of operations, from the inner parentheses outward.

According to the rule, when it is in a negative power, it is made into a positive by replacing the numerator and denominator.

17βˆ’1=71 \frac{1}{7^{-1}}=7^1

Now, we use the power law to multiply the power that is enclosed in parentheses with the one inside.

1Γ—4=4 1\times4=4

Thus we arrive at the solution

Answer:

74 7^4


Do you think you will be able to solve it?

Exercise 4

Solve the following exercise:

1X7X6= \frac{1}{\frac{X^7}{X^6}}=

Solution:

First we note the fraction in the denominator of the exercise.

Also here two laws are used, first of all the power law, according to which we do

x7x6=x(7βˆ’6)=x1=x \frac{x^7}{x^6}=x^{(7-6)}=x^1=x

Now, we subtract only the fraction

1x \frac{1}{x}

We know that this form can also be converted through the property of the negative exponents, so we can also write:

Answer:

1x=xβˆ’1 \frac{1}{x}=x^{^{-1}}


Exercise 5

Simplify the expression (x2(2y)βˆ’3)βˆ’2 \left(\frac{x^2}{\left(2y\right)^{-3}}\right)^{-2}

Solution

First we solve the expression inside the parentheses. The denominator has a negative exponent so we can place it in the numerator and change the sign of the exponent.

(x2(2y)βˆ’3)βˆ’2=(x2(2y)3)βˆ’2 \left(\frac{x^2}{\left(2y\right)^{-3}}\right)^{-2}=\left(x^2\left(2y\right)^3\right)^{-2}

We raise 2y 2y to the indicated power and reorder the expression.

(x2(2y)3)βˆ’2=(x2β‹…8y3)βˆ’2=(8x2y3)βˆ’2 \left(x^2\left(2y\right)^3\right)^{-2}=\left(x^2\cdot8y^3\right)^{-2}=\left(8x^2y^3\right)^{-2}

Finally we apply the rule of negative exponents.

(8x2y3)βˆ’2=1(8x2y3)2=164x4y6 \left(8x^2y^3\right)^{-2}=\frac{1}{\left(8x^2y^3\right)^{2}}=\frac{1}{64x^4y^6}

Answer:

(x2(2y)βˆ’3)βˆ’2=164x4y6 \left(\frac{x^2}{\left(2y\right)^{-3}}\right)^{-2}= \frac{1}{64x^4y^6}


Test your knowledge

Review questions

What happens when there is a power with a negative exponent?

When we have a power with a negative exponent, we can make the exponent positive by converting the expression to a fraction. First we put a 1 in the numerator, in the denominator we put the original power, but changing the negative sign of the exponent by a positive sign.


When the exponent is negative, is the result is negative?

The negative sign of the exponent does not mean that the result can't be positive. Look at the following example

(3)βˆ’1=13 \left(3\right)^{-1}=\frac{1}{3}

Notice how the sign of the exponent is negative but the result is 1/3 positive.


How do you proceed when the exponent is negative or positive?

If the exponent is positive we just multiply the base by itself as many times as the exponent indicates. If the exponent is negative we first convert the expression to a fraction with positive exponents and proceed as mentioned above.


What is done when the base of a power is negative?

If the base of a power is negative and the exponent is an integer, we are only indicating the sign of the number that will be multiplied by itself, as many times as the exponent indicates. If the exponent is an even number, the sign of the power will be positive, but if it is odd, the sign of the power will be negative.


What happens when the base is negative and the exponent is 0 0 ?

If the base is a negative number with zero exponent the result is 1 1 .


Exercises for negative exponents

Exercise 1

Task

7βˆ’24=? 7^{-24}=\text{?}

Solution

7βˆ’24=70βˆ’24= 7^{-24}=7^{0-24}=

70724=1724 \frac{7^0}{7^{24}}=\frac{1}{7^{24}}

Answer

1724 \frac{1}{7^{24}}


Do you know what the answer is?

Exercise 2

Task

(8Γ—9Γ—5Γ—3)βˆ’2= (8\times9\times5\times3)^{-2}=

Solution

We will use the formula

(abc)n=anβ‹…bnβ‹…cn \left(abc\right)^n=a^n\cdot b^n\cdot c^n

8βˆ’2β‹…9βˆ’2β‹…5βˆ’2β‹…3βˆ’2 8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}

Answer

8βˆ’2Γ—9βˆ’2Γ—5βˆ’2Γ—3βˆ’2 8^{-2}\times9^{-2}\times5^{-2}\times3^{-2}


Exercise 3

Task

((7Γ—3)2)6+(3βˆ’1)3Γ—(23)4= ((7\times3)^2)^6+(3^{-1})^3\times(2^3)^4=

Solution

We will use the formula

(am)n=amβ‹…n \left(a^m\right)^n=a^{m\cdot n}

(7β‹…3)2β‹…6+3βˆ’1β‹…3β‹…23β‹…4= \left(7\cdot3\right)^{2\cdot6}+3^{-1\cdot3}\cdot2^{3\cdot4}=

2112+3βˆ’3β‹…212 21^12+3^{-3}\cdot2^{12}

Answer

2112+3βˆ’3Γ—212 21^{12}+3^{-3}\times2^{12}


Check your understanding

Exercise 4

Task

(3Γ—2Γ—4Γ—6)βˆ’4= (3\times2\times4\times6)^{-4}=

Solution

We will use the formula

(aβ‹…bβ‹…c)n=anβ‹…bnβ‹…cn \left(a\cdot b\cdot c\right)^n=a^n\cdot b^n\cdot c^n

(3β‹…2β‹…4β‹…6)βˆ’4=3βˆ’4β‹…2βˆ’4β‹…4βˆ’4β‹…6βˆ’4 \left(3\cdot2\cdot4\cdot6\right)^{-4}=3^{-4}\cdot2^{-4}\cdot4^{-4}\cdot6^{-4}

Answer

3βˆ’4Γ—2βˆ’4Γ—4βˆ’4Γ—6βˆ’4 3^{-4}\times2^{-4}\times4^{-4}\times6^{-4}


Exercise 5

Task

19βˆ’2=? 19^{-2}=\text{?}

Solution

19βˆ’2=190βˆ’2 19^{-2}=19^{0-2}

190192=1192= \frac{19^0}{19^2}=\frac{1}{19^2}=

1361 \frac{1}{361}

Answer

1361 \frac{1}{361}


Do you think you will be able to solve it?
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