CBSE Class 10 Social Science

May 24, 202339 Insightful Publications

**Powers with Negative Exponents:** We are not convenient to read, understand and compare large numbers like \(75,00,00,000;1,459,500,000,000;5,978,043,000,000,000;\) etc. To make such large numbers easy to read, understand and compare, we use exponents. We can write \(6 \times 6 \times 6 \times 6\) as \({6^4}\) and it is read as \(6\) raised to the power \(4.\) In \({6^4},\) we call \(6\) as the base and \(4\) the exponent.

This notation is called exponential form or power notation. We know that \({10^2} = 10 \times 10 = 100,{10^1} = 10 = \frac{{100}}{{10}},{10^0} = 1 = \frac{{10}}{{10}},{10^{\left({ – 1} \right)}} = \frac{1}{{10}}\) Here, \( – 1\) is the negative exponent of the base \(10.\) As the exponent decreases by \(1,\) the value becomes one-tenth of the previous value. In this article, we will learn about the powers with negative exponents, their properties, and problems based on the negative exponents.

The method of writing large numbers in a shorter form using the powers is known as exponential form.

Example: \(10000 = 10 \times 10 \times 10 \times 10 = {10^4}\)

The short notation \({10^4}\) stands for the product \(10 \times 10 \times 10 \times 10.\) Here \(10\) is called the base, and \(4\) is called an exponent.

**Learn the Concepts of Exponents and Powers**

We have used numbers like \({\text{10,100,1000,}}\) etc., while writing numbers in an expanded form.

Example:\(58761 = 5 \times 10000 + 8 \times 1000 + 7 \times 100 + 6 \times 10 + 1\)

Above expansion can be written as \(5 \times {10^4} + 8 \times {10^3} + 7 \times {10^2} + 6 \times {10^1} + 1\)

In the above example, we have seen numbers whose base is \(10.\) However, the base can be any other number also.

Example: \(16 = 2 \times 2 \times 2 \times 2 = {2^4}.\) Here, \(2\) is the base and \(4\) is the exponent.

Let us consider some of the examples: \({2^5} \times {2^3},{3^2} \times {3^4},\frac{{{4^7}}}{{{4^4}}},{\left({{2^3}} \right)^2}\)

To find out the values of the above examples, we have some laws of exponents. Let us study those laws in the next section.

In this section, we will learn about various laws of exponents.

If \(a\) is any non-zero rational number and \(m,n\) are natural numbers, then \({a^m} \times{a^n} = {a^{\left({m + n} \right)}}\)

Also, if a is any non-zero rational number and \(m,n,p\) are natural numbers, then \({a^m} \times{a^n} \times {a^p} = {a^{\left({m + n + p} \right)}}\)

Example: \({3^2} \times {3^4} = \left({3 \times 3} \right) \times \left({3 \times 3 \times 3 \times 3} \right) = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = {3^6} = {3^{\left({2 + 4} \right)}}\)

Example: \(\frac{{{4^7}}}{{{4^4}}} = \frac{{4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4}}{{4 \times 4 \times 4 \times 4}} = 4 \times 4 \times 4 = {4^3} = {4^{\left({7 – 4} \right)}}\)

Example: \({\left({{4^2}} \right)^3} = {4^2} \times {4^2} \times {4^2} = {4^{\left({2 + 2 + 2}\right)}} = {4^6} = {4^{\left({2 \times 3} \right)}}\)

Example: \({3^4} \times {5^4} = \left( {3 \times 3 \times 3 \times 3} \right) \times \left( {5 \times 5 \times 5 \times 5} \right)\)

\( = \left( {3 \times 5} \right) \times \left( {3 \times 5} \right) \times \left( {3 \times 5} \right) \times \left( {3 \times 5} \right)\)

\( = 15 \times 15 \times 15 \times 15 = {15^4} = {\left( {3 \times 5} \right)^4}\)

Example: \(\frac{{{3^3}}}{{{5^3}}} = \frac{{3 \times 3 \times 3}}{{5 \times 5 \times 5}} = \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} = {\left({\frac{3}{5}} \right)^3}\)

Let us now find the value of a power of a non-zero rational number when its exponent is zero.

We know that \(\frac{{{5^4}}}{{{5^4}}} = \frac{{5 \times 5 \times 5 \times 5}}{{5 \times 5 \times 5 \times 5}} = 1…….\left( i \right)\)

Also, \(\frac{{{5^4}}}{{{5^4}}} = {5^{(4 – 4)}} = {5^ \circ }…….\left( {ii} \right)\)

From \(\left( i \right)\) and \(\left( ii \right),\) we can write as \(\frac{{{5^4}}}{{{5^4}}} ={5^{(4 – 4)}} = {5^ \circ } \ldots \ldots \left({ii} \right)\)

Therefore, for any non-zero rational number \(a\) we have \({a^ \circ } = 1.\)

In the above section, we have learnt that

\({10^ \circ } = 1\)

\({10^1} = 10\)

\({10^2} = 100\)

\({10^3} = 1000\) and so on.

Also, we know that

\(\frac{{10000}}{{10}} = 1000\)

\(\frac{{1000}}{{10}} = 100\)

\(\frac{{100}}{{10}} = 10\)

\(\frac{{10}}{{10}} = 1\)

In exponential form, the above results can be written as follows:

\(\frac{{{{10}^4}}}{{10}} ={10^3}\) or \({10^3}=\frac{{{{10}^4}}}{{10}}\)

\(\frac{{{{10}^3}}}{{10}} ={10^2}\) or \({10^2}=\frac{{{{10}^3}}}{{10}}\)

\(\frac{{{{10}^2}}}{{10}} ={10^1}\) or \({10^1}=\frac{{{{10}^2}}}{{10}}\)

\(\frac{{{{10}^1}}}{{10}} = 1 = {10^ \circ }\) or \({10^ \circ } = 1\frac{{{{10}^1}}}{{10}}\)

The above results exhibit a pattern that, as the exponent of \(10\) decreases by \(1,\) the value becomes one-tenth of the previous value. So, if the same pattern is continued, we must have

\({10^{\left({ – 1} \right)}} = \frac{1}{{10}}\)

\({10^{\left({ – 2} \right)}} = \frac{1}{{10}} \div 10 = \frac{1}{{10}} \times \frac{1}{{10}} = \frac{1}{{100}}\)

\({10^{\left({ – 3} \right)}} = \frac{1}{{100}} \div 10 = \frac{1}{{100}} \times \frac{1}{{10}} = \frac{1}{{1000}}\)

\({10^{\left({ – 4} \right)}} = \frac{1}{{1000}} \div 10 = \frac{1}{{1000}} \times \frac{1}{{10}} = \frac{1}{{10000}}\)

The above results suggest the following definition for the negative-integral exponents of a non-zero rational number.

For any non-zero rational number \(a\) and a positive integer, we define\({a^{\left({ – n} \right)}} = \frac{1}{{{a^n}}}i.e.,{a^{\left({ – n} \right)}}\) is the reciprocal of \({a^n}.\)

Hence, \({\left( {{2^3}} \right)^2} = {2^{\left( {3 \times 2} \right)}} = {2^6}.\) So, \({\left( {{2^{ – 3}}} \right)^{ – 2}} = {2^{\left( { – 3 \times 2} \right)}} = {2^6}.\)

Now, let us see whether the above laws also hold if the exponents are negative?

The first law states that, if \(a\) is any non-zero rational number and \(m,n\) are natural numbers, then \({a^m} \times {a^n} = {a^{\left({m + n} \right)}}\)

We know that \({3^{\left({ – 3} \right)}} = \frac{1}{{{3^3}}}\) and \({3^{\left({ – 2} \right)}} = \frac{1}{{{3^2}}}\)

Therefore, \({3^{\left({ – 3} \right)}} \times {3^{\left({ – 2} \right)}} = \frac{1}{{{3^3}}} \times \frac{1}{{{3^2}}} = \frac{1}{{{3^3} \times {3^2}}} = \frac{1}{{{3^{\left({3 + 2} \right)}}}} = \frac{1}{{{3^5}}} = {3^{\left({ – 5} \right)}}\)

Clearly, \( – 5\) is the sum of \( – 3\) and \( – 2.\) So, first law, i.e., \({a^m} \times {a^n} = {a^{\left({m + n} \right)}}\) holds for the negative exponents. In other words, \({a^{ – m}} \times {a^{ – n}} = {a^{ – \left({m + n} \right)}}\)

The second law states that, if \(a\) is any non-zero rational number and \(m,n\) are natural numbers such that \(m > n,\) then \({a^m} \div {a^n} = {a^{\left({m – n}\right)}}\) or \(\frac{{{a^m}}}{{{a^n}}} = {a^{\left({m – n} \right)}}\)

Now, consider \({2^{\left({ – 3} \right)}}\) and \({2^{\left({ – 2} \right)}}\)

\({2^{\left({ – 3} \right)}} \div {2^{\left({ – 2} \right)}} = \frac{1}{{{2^3}}} \div \frac{1}{{{2^2}}} = \frac{1}{{{2^3}}} \times \frac{{{2^2}}}{1} = \frac{{{2^2}}}{{{2^3}}} = {2^{\left({2 – 3} \right)}} = {2^{\left({ – 1} \right)}}\)

So, it is clear that \(\left({2 – 3} \right) = \, – 1.\) This means second law, i.e., \({a^m} \div {a^n} = {a^{\left({m – n} \right)}}\) holds for the negative exponents. In other words, \({a^{ – m}} \div {a^{ – n}} = {a^{\left({ – m + n} \right)}}\)

The third law states that, If \(a\) is any non-zero rational number and \(m,n\) are natural numbers, then \({\left({{a^m}}\right)^n} = {a^{\left({m \times n}\right)}} = {\left({{a^n}} \right)^m}\)

Hence, \({\left({{2^3}} \right)^2} = {2^{\left({3 \times 2} \right)}} = {2^6}.\) So, \({\left({{2^{ – 3}}} \right)^{ – 2}} = {2^{\left({ – 3x – 2} \right)}} = {2^6}\)

Therefore, the third law, i.e., \({\left({{a^m}} \right)^n} = {a^{\left({m \times n} \right)}} = {\left({{a^n}} \right)^m}\) holds suitable for negative exponents. We can write it as \({\left({{a^{ – m}}} \right)^{ – n}} = {a^{\left({m \times n} \right)}}\)

The fourth law states that, if \(a,b\) are non-zero rational numbers and \(n\) is a natural number, then \({a^n} \times {b^n} = {\left({ab} \right)^n}\)

Now consider, \({2^{\left({ – 3}\right)}} \times {3^{\left({ – 3} \right)}} = \frac{1}{{{2^3}}} \times \frac{1}{{{3^3}}} = \frac{1}{{{{\left({2 \times 3} \right)}^3}}} = {\left({2 \times 3}\right)^ – }3\)

Therefore, the fourth law, i.e., \({a^n} \times {b^n} = {\left({ab} \right)^n}\) holds suitable for negative exponents. We can write it as \({a^{ – n}} \times {b^{ – n}} = {\left({ab} \right)^{ – n}}\)

The fifth law states that, if \(a\) and \(b\) are non-zero rational numbers and \(n\) is a natural number, then \(\frac{{{a^n}}}{{{b^n}}} = {\left({\frac{a}{b}} \right)^n}\)

Consider, \(\frac{{{4^{ – 3}}}}{{{5^{ – 3}}}} = \frac{{{5^3}}}{{{4^3}}} = \frac{{5 \times 5 \times 5}}{{4 \times 4 \times 4}} = {\left({\frac{5}{4}} \right)^3} ={\left({\frac{4}{5}} \right)^{ – 3}}\)

Therefore, the fifth law, i.e.,\(\frac{{{a^n}}}{{{b^n}}} = {\left({\frac{a}{b}} \right)^n}\) holds good for the negative exponents. So, \(\frac{{{a^{ – n}}}}{{{b^{ – n}}}} = {\left({\frac{a}{b}} \right)^{ – n}}\)

We know how to express very large numbers in standard form by using exponents of \(10.\) Let us study how can we write very small numbers in standard form, also known as scientific notation, by using the following steps:

(i) Obtain the number and see whether the number is between \(1\) and \(10\) or less than \(1.\)

(ii) If the number is between \(1\) and \(10,\) then write it as the product of the number itself and \({10^ \circ }.\)

(iii) If the number is less than one, then move the decimal point to the right to just one digit on the left side of the decimal point. Write the given number as the product of the number so obtained and \({10^{ – n}},\) where \(n\) is the number of places by which the decimal point has been moved to the right. The number so obtained is the standard form of the given number.

**Q.1. By what number should \({\left( { – 24} \right)^{ – 1}}\) be divided so that we get \({3^{ – 1}}?\) ****Ans: **Let the required number be \(x.\) Then, \({\left({ – 24} \right)^{ – 1}} \div x = {3^{ – 1}}\)

\( \Rightarrow \frac{{{{\left({ – 24} \right)}^{ – 1}}}}{x} = {3^{ – 1}}\)

\( \Rightarrow \frac{{\frac{1}{{ – 24}}}}{x} = \frac{1}{3}\) (Because \({a^{ – 1}} = \frac{1}{a}\))

\( \Rightarrow \frac{1}{{ – 24x}} = \frac{1}{3}\)

\( \Rightarrow 3 = – 24x\)

\(\Rightarrow x = \frac{3}{{ – 24}}\)

\( \Rightarrow x = – \frac{1}{8}\)

So, we need to divide \({( – 24)^{ – 1}}\) by \( – \frac{1}{8}\) to get the quotient \({3^{ – 1}}.\)

**Q.2. By what number should \({\left( { – 8} \right)^{ – 1}}\) be multiplied so that the product is equal to \({10^{ – 1}}?\)Ans:** Let \({\left({ – 8} \right)^{ – 1}}\) be multiplied by \(x\) to get \({10^{ – 1}}.\) Then,

\(x \times {\left({ – 8} \right)^{ – 1}} = {10^{ – 1}}\)

\( \Rightarrow x = {10^{ – 1}} \div {\left({ – 8} \right)^{ – 1}}\)

\( \Rightarrow x = \frac{1}{{10}} \div \frac{1}{8}\) (Because \({a^{ – 1}} = \frac{1}{a}\))

\( \Rightarrow x = \frac{1}{{10}} \times \frac{{ – 8}}{1} = \frac{{ – 8}}{{10}} = \frac{{ – 4}}{5}\)

Hence, we need to multiply \( – \frac{4}{5}\) with \({\left({ – 8} \right)^{ – 1}}\) to get \({10^{ – 1}}.\)

**Q.3. Express each of the following as a rational number of the form \(\frac{p}{q}.\)(i) \({\left( {{2^{ – 1}} + {3^{ – 1}}} \right)^2}\)(ii) \({\left( {{2^{ – 1}} – {4^{ – 1}}} \right)^2}\)(iii) \(\left\{ {{{\left( {\frac{3}{4}} \right)}^{ – 1}} – {{\left( {\frac{1}{4}} \right)}^{ – 1}}} \right\}\)Ans: **We know that for any positive integer \(n\) and any rational number \(a,{a^{ – n}} = \frac{1}{{{a^n}}}\)

Therefore, we have

(i) \({\left({{2^{ – 1}} + {3^{ – 1}}} \right)^2} = {\left({\frac{1}{2} + \frac{1}{3}} \right)^2} = {\left({\frac{{3 + 2}}{6}} \right)^2} = {\left({\frac{5}{6}} \right)^2} = \frac{{{5^2}}}{{{6^2}}} = \frac{{25}}{{36}}\)

Therefore, \({\left({{2^{ – 1}} + {3^{ – 1}}} \right)^2} = \frac{{25}}{{36}}\)

(ii) \({\left({{2^{ – 1}} – {4^{ – 1}}} \right)^2}\) \( = {\left({\frac{1}{2} – \frac{1}{4}} \right)^2}\) (Because \({a^{ – 1}} = \frac{1}{a}\))

\( = {\left({\frac{{2 – 1}}{4}} \right)^2}\)

\( = {\left({\frac{1}{4}} \right)^2}\)

\( = \frac{{{1^2}}}{{{4^2}}}\)(Because \(\frac{{{a^n}}}{{{b^n}}} = {\left({\frac{a}{b}} \right)^n}\))

\( = \frac{1}{{16}}\)

Therefore, \({\left({{2^{ – 1}} – {4^{ – 1}}} \right)^2} = \frac{1}{{16}}\)

(iii) \(\left\{{{{\left({\frac{3}{4}} \right)}^{ – 1}} – {{\left( {\frac{1}{4}} \right)}^{ – 1}}} \right\}\)

\( = {\left({\frac{1}{{\frac{3}{4}}} – \frac{1}{{\frac{1}{4}}}} \right)^{ – 1}}\)(Because \({a^{ – 1}} = \frac{1}{a}\))

\( = {\left({\frac{4}{3} – \frac{4}{1}} \right)^{ – 1}}\)

\( = {\left({\frac{{4 – 12}}{3}} \right)^{ – 1}}\)

\( = {\left({\frac{{ – 8}}{3}} \right)^{ – 1}}\)

\( = \frac{1}{{\frac{{ – 8}}{3}}}\) (Because \({a^{ – 1}} = \frac{1}{a}\))

\( = \frac{{ – 3}}{8}\)

Therefore, \(\left\{{{{\left({\frac{3}{4}} \right)}^{ – 1}} – {{\left({\frac{1}{4}} \right)}^{ – 1}}} \right\} = \frac{{ – 3}}{8}\)

**Q.4. Simplify:\({\left( {\frac{1}{4}} \right)^{ – 2}} + {\left( {\frac{1}{2}} \right)^{ – 2}} + {\left( {\frac{1}{3}} \right)^{ – 2}}\)\({\left\{ {{6^{ – 1}} + {{\left( {\frac{3}{2}} \right)}^{ – 1}}} \right\}^{ – 1}}\)Ans:** We have,

(i) \({\left({\frac{1}{4}}\right)^{ – 2}} + {\left({\frac{1}{2}} \right)^{ – 2}} + {\left({\frac{1}{3}} \right)^{ – 2}}\)

\( = \frac{1}{{{{\left({\frac{1}{4}} \right)}^2}}} + \frac{1}{{{{\left( {\frac{1}{2}} \right)}^2}}} + \frac{1}{{{{\left({\frac{1}{3}} \right)}^2}}}\) (Because \({a^{ – n}} = \frac{1}{{{a^n}}}\))

\( = \frac{1}{{\frac{{{1^2}}}{{{4^2}}}}} + \frac{1}{{\frac{{{1^2}}}{{{2^2}}}}} + \frac{1}{{\frac{{{1^2}}}{{{3^2}}}}}\) (Because \(\frac{{{a^n}}}{{{b^n}}} = {\left({\frac{a}{b}} \right)^n}\))

\( = \frac{{{4^2}}}{{{1^2}}} + \frac{{{2^2}}}{{{1^2}}} + \frac{{{3^2}}}{{{1^2}}}\)

\( = {4^2} + {2^2} + {3^2}\)

\( = 16 + 4 + 9 = 29\)

Therefore, \({\left({\frac{1}{4}} \right)^{ – 2}} + {\left({\frac{1}{2}} \right)^{ – 2}} + {\left({\frac{1}{3}} \right)^{ – 2}} = 29\)

(ii) \({\left\{{{6^{ – 1}} + {{\left({\frac{3}{2}} \right)}^{ – 1}}} \right\}^{ – 1}}\)

\( = {\left\{{\frac{1}{6} + \frac{1}{{\frac{3}{2}}}} \right\}^{ – 1}}\) (Because \({a^{ – 1}} = \frac{1}{a}\))

\( = {\left\{{\frac{1}{6} + \frac{2}{3}} \right\}^{ – 1}}\)

\( = {\left\{{\frac{{1 + 4}}{6}} \right\}^{ – 1}}\)

\( = {\left\{{\frac{5}{6}}\right\}^{ – 1}}\)

\( = \frac{1}{{\frac{5}{6}}}\)

\( = \frac{6}{5}\)

Therefore, \({\left\{{{6^{ – 1}} + {{\left({\frac{3}{2}} \right)}^{ – 1}}} \right\}^{ – 1}} = \frac{6}{5}\)

**Q.5. Simplify \({\left( {{2^{ – 1}} \div {5^{ – 1}}} \right)^2} \times {\left( {\frac{{ – 5}}{8}} \right)^{ – 1}}\) Ans:** We have \({\left({{2^{ – 1}} \div {5^{ – 1}}} \right)^2} \times {\left({\frac{{ – 5}}{8}} \right)^{ – 1}}\)

\( = {\left({\frac{1}{2} \div \frac{1}{5}} \right)^2} \times \frac{1}{{\frac{{ – 5}}{8}}}\) (Because \({a^{ – 1}} = \frac{1}{a}\))

\( = {\left({\frac{1}{2} \times \frac{5}{1}} \right)^2} \times \frac{8}{{ – 5}}\)

\( = {\left({\frac{5}{2}} \right)^2} \times \frac{8}{{ – 5}}\)

\( = \frac{{{5^2}}}{{{2^2}}} \times \frac{8}{{ – 5}}\) (Because \(\frac{{{a^n}}}{{{b^n}}} = {\left({\frac{a}{b}} \right)^n}\))

\( = \frac{5}{4} \times \frac{8}{{ – 1}}\)

\( = \frac{5}{1} \times \frac{2}{{ – 1}}\)

\( = – 10\)

Therefore, \({\left({{2^{ – 1}} \div {5^{ – 1}}} \right)^2} \times {\left({\frac{{ – 5}}{8}} \right)^{ – 1}} = \,- 10\)

**Q.6. Write the number 0.4579 in the standard form. ****Ans:** To express \(0.4579\) in standard form, the decimal point is moved through one place only to the right so that there is just one digit on the left of the decimal point.

So, \(0.4579 = 4.579 \times {10^{ – 1}}\)

In the above article, we have learnt the definition of exponents, different laws of exponents and power with exponent zero. Also, we have learnt the meaning of power with negative exponents and the power rules with negative exponents and solved some example problems on negative exponents.

**Q.1. Where do we use negative exponents?Ans: **A positive exponent tells us how many times we need to multiply a base number, and a negative exponent tells us how many times we need to divide a base number. Also, we can write the negative exponent as \({a^{ – n}} = \frac{1}{{{a^n}}}\)

**Q.2. What happens when we rise to a negative power?****Ans:** Raising a number to a negative exponent isn’t much different from raising a positive exponent. The only difference is that a negative exponent makes you take the reciprocal of the base first. This will turn the expression into one with a positive exponent.

**Q.3. How to write numbers in scientific notation?Ans:** We can write very small numbers in standard form, also known as scientific notation, by using the following steps:

(i) Obtain the number and see whether the number is between \(1\) and \(10\) or less than \(1.\)

(ii) If the number is between \(1\) and \(10,\) then write it as the product of the number itself and \({10^ \circ }.\)

(iii) If the number is less than one, then move the decimal point to the right to just one digit on the left side of the decimal point. Write the given number as the product of the number so obtained and \({10^{ – n}}\) where \(n\) is the number of places by which the decimal point has been moved to the right. The number so obtained is the standard form of the given number.

**Q.4. What does 10 to the power negative 2 mean? ****Ans: **\({10^{ – 2}} = \frac{1}{{{{10}^2}}} = \frac{1}{{100}} = 0.01\)

**Q.5. How do we know if the exponent is negative in scientific notation?Ans:** If we have the smaller number in the decimal form, i.e., smaller than \(1.\) Then, the power is negative.