Powers with Negative Exponents

Large numbers like \[75,000,000\], \[1,459,500,000,000\], \[5,978,043,000,000,000\], etc. are difficult for us to read, understand, and compare. We take the help of exponents to make such large numbers simple to read, understand, and compare. For example, the number \[6 \times 6 \times 6 \times 6\] is understood as $6$ raised to the power $4$. In the number ${6^4}$, the exponent is $4$ and the base is $6$.

This notation is also known as power notation or exponential form. Since \[{10^2} = 10 \times 10 = 100\], \[{10^1} = 10 = \dfrac{{100}}{{10}}\], \[{10^0} = 1 = \dfrac{{10}}{{10}}\], and \[{10^{ - 1}} = \dfrac{1}{{10}}\], we know that. The base $10$’s negative exponent, $ - 1$, is used here. The value becomes one-tenth of the previous value when the exponent reduces by $1$. The powers with negative exponents, their characteristics, and the problems based on them will all be covered in this article.

What are Exponents?

Exponential form refers to the use of powers to reduce large numbers into a more simplified form.

An example is \[10000 = 10 \times 10 \times 10 \times 10 = {10^4}\] .

\[10 \times 10 \times 10 \times 10\] is represented by the abbreviation ${10^4}$ . In this case, $10$ serves as the base and $4$ as the exponent.

Rules of Negative Exponents

We have a series of principles or rules for negative exponents which make the process of simplification easy. The fundamental guidelines for resolving negative exponents are listed below.

Rule 1: The reciprocal of a base, which is $\dfrac{1}{a}$, is multiplied by itself $n$ times according to the negative exponent rule for bases with the negative exponent $ - n$.

Specifically, ${a^{( - n)}} = \dfrac{1}{a} \times \dfrac{1}{a} \times \ldots n$ times which equals $\dfrac{1}{{{a^n}}}$.

Rule 2: The rule holds true even when the denominator contains a negative exponent.

In other words, $\dfrac{1}{{{a^{( - n)}}}} = a \times a \times \ldots n$ which is equal to ${a^n}$ .

Negative Laws of Exponents

• According to the first law, if $m$ and $n$ are natural numbers and $a$ is any non-zero rational integer, then ${a^m} \times {a^n} = {a^{m + n}}$ .

${3^{ - 3}} = \dfrac{1}{{{3^3}}}$ and ${3^{ - 2}} = \dfrac{1}{{{3^2}}}$ are known numbers.

In light of this, ${3^{ - 3}} \times {3^{ - 2}} = \dfrac{1}{{{3^3}}} \times \dfrac{1}{{{3^2}}} = \dfrac{1}{{{3^{3 + 2}}}} = \dfrac{1}{{{3^5}}} = {3^{ - 5}}$ .

Clearly, adding $ - 3$ and $ - 2$ results in $ - 5$ . Therefore, the first law, ${a^m} \times {a^n} = {a^{m + n}}$, still applies to negative exponents. Alternatively, ${a^{ - m}} \times {a^{ - n}} = {a^{ - (m + n)}}$ .

• According to the second law, if $a$ is any non-zero rational number and $m$, $n$ are natural numbers with $m > n$, then $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ or ${a^m} \div {a^n} = {a^{m - n}}$.

Now, think about ${2^{ - 3}}$ and ${2^{ - 2}}$.

${2^{ - 3}} \div {2^{ - 2}} = \dfrac{1}{{{2^3}}} \div \dfrac{1}{{{2^2}}} = \dfrac{1}{{{2^3}}} \times \dfrac{{{2^2}}}{1} = {2^{2 - 3}} = {2^{ - 1}}$ .

Therefore, it is evident that $(2 - 3) = - 1$. This suggests that the second law, ${a^m} \div {a^n} = {a^{m - n}}$, is valid for negative exponents. Alternatively, ${a^{ - m}} \div {a^{ - n}} = {a^{ - (m - n)}}$.

• The third law asserts that ${({a^m})^n} = {a^{m \times n}} = {({a^n})^m}$ if $a$ is any non-zero rational integer and $m$, $n$ are natural numbers.

Hence, \[{({2^3})^2} = {2^{3 \times 2}} = {2^6}\] . So, \[{({2^{ - 3}})^{ - 2}} = {2^{( - 3) \times ( - 2)}} = {2^6}\].

As a result, the third law, ${({a^m})^n} = {a^{m \times n}} = {({a^n})^m}$, is valid for negative exponents. The formula is ${({a^{ - m}})^{ - n}} = {a^{( - m) \times ( - n)}} = {({a^{ - n}})^{ - m}}$.

• According to the fourth law, if $a$ and $b$ are rational non-zero numbers and $n$ is a natural number, then ${a^n} \times {b^n} = {(ab)^n}$ .

Consider the following:

${2^{ - 3}} \times {3^{ - 3}} = \dfrac{1}{{{2^3}}} \times \dfrac{1}{{{3^3}}} = \dfrac{1}{{{{(2 \times 3)}^3}}} = {(2 \times 3)^{ - 3}}$

Because of this, the fourth law, ${a^n} \times {b^n} = {(ab)^n}$, still applies to negative exponents. It can be written as ${a^{ - n}} \times {b^{ - n}} = {(ab)^{ - n}}$.

• If $a$ and $b$ are non-zero rational numbers and $n$ is a natural number, the fifth law states that $\dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n}$.

Consider the expression $\dfrac{{{4^{ - 3}}}}{{{5^{ - 3}}}} = \dfrac{{{5^3}}}{{{4^3}}} = \dfrac{{5 \times 5 \times 5}}{{4 \times 4 \times 4}} = {\left( {\dfrac{5}{4}} \right)^3} = {\left( {\dfrac{4}{5}} \right)^{ - 3}}$ .

As a result, $\dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n}$, the fifth law still applies to negative exponents. Hence, $\dfrac{{{a^{ - n}}}}{{{b^{ - n}}}} = {\left( {\dfrac{a}{b}} \right)^{ - n}}$ .

How to Solve Negative Exponents

Use one of the following rules to change negative exponents into positive exponents before simplifying equations with negative exponents:

• \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]

• \[\dfrac{1}{{{a^{ - n}}}} = {a^n}\]

Negative Exponents Examples

1. Find the solution to the expression ${({3^2} + {4^2})^{ - 2}}$.

Solution: The provided expression is

\[{\left( {{3^2} + {4^2}} \right)^{ - 2}} = {\left( {9 + 16} \right)^{ - 2}}\]

\[ = {\left( {25} \right)^{ - 2}}\]

\[ = \dfrac{1}{{{{25}^2}}}\] (by negative exponents rule)

$ = \dfrac{1}{{625}}$

Because of this, ${({3^2} + {4^2})^{ - 2}} = \dfrac{1}{{625}}$.

2. Condense ${({2^{ - 1}} \div {5^1})^2} \times {\left( {\dfrac{{ - 5}}{8}} \right)^{ - 1}}$ .

Solution: The number is ${({2^{ - 1}} \div {5^1})^2} \times {\left( {\dfrac{{ - 5}}{8}} \right)^{ - 1}}$

because ${a^{ - 1}} = \dfrac{1}{a}$ $ \Rightarrow {\left( {\dfrac{1}{2} \div \dfrac{1}{5}} \right)^2} \times \dfrac{1}{{\dfrac{{ - 5}}{8}}}$

$ = {\left( {\dfrac{1}{2} \times \dfrac{5}{1}} \right)^2} \times \dfrac{8}{{ - 5}}$

$ = {\left( {\dfrac{5}{2}} \right)^2} \times \dfrac{8}{{ - 5}}$

Because $\dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n}$

$ = \dfrac{5}{4} \times \dfrac{8}{{ - 1}}$

$ = \dfrac{5}{1} \times \dfrac{2}{{ - 1}}$

\[ = - 10\]

Practice Questions

• Using the principles for negative exponents, find the unknown $x$ in the equation ${5^x} = \dfrac{1}{{625}}$

• When \[\dfrac{{27}}{{{3^{ - x}}}} = {3^6}\] , what is the value of $x$?

Answers

• $ - 4$

• $x = 3$

Summary

The definition of exponents, various negative exponent laws, and power with exponent negative were all covered in the article above. Additionally, we have mastered the definition of power with negative exponents, and the power laws with negative exponents, and have worked through a few practice problems on negative exponents.