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ERR - 7th - Maths [Exponents and Power II]

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7th (Maths)
Exponents and Power II
Law 4 Power of Exponents If we have to take the power of power then we have to multiply the exponents i.e. $(a^{m})^{n}=a^{mn}$ If we have to divide the powers where the base is different but exponents are the same then we will divide the base. i.e. $left ( frac{a^{m}}{b^{m}} right )=left ( frac{a}{b} right )^{m}$ Illustration: Evaluate $(3^{3})^{5}$ Solution: The given expression is $(3^{3})^{5}$ As we know that $(a^{m})^{n}=a^{mn}$ Hence, $(3^{3})^{5}=(3)^{15}$ Illustration: Evaluate the expression given below $left ( frac{2^{6}}{5^{6}} right )$ . Solution: The given expression is $left ( frac{2^{6}}{5^{6}} right )$ As we know that $left ( frac{a^{m}}{b^{m}} right )=left ( frac{a}{b} right )^{m}$ Hence, we have $left ( frac{2^{6}}{5^{6}} right )=left ( frac{2}{5} right )^{6}$ Illustration: Simplify the expression given below $left ( frac{left ( frac{2}{5}right )^{6}}{7^{6}} right )$ Solution: $left ( frac{left ( frac{2}{5}right )^{6}}{7^{6}} right )=left ( frac{2}{5} right )^{6}times left ( frac{1}{7} right )^{6}$ As we know that $left ( frac{a^{m}}{b^{m}} right )=left ( frac{a}{b} right )^{m}$ Hence, we have $=left ( frac{2times 1}{5times 7} right )^{6}$ $=left ( frac{2}{35} right )^{6}$
Law 5 Numbers with Zero Exponent Any number with zero exponents is equal to one irrespective of the base i.e., $a^{0}=1$ Any number with one as the exponent is equal to the number itself i.e., $a^{1}=a$ Examples: 1. $8^{0}=1$ 2. $(-12)^{0}=1$ 3. $left ( frac{5}{9} right )^{0}=1$ 4. $left ( frac{-7}{11} right )^{1}=left ( frac{-7}{11} right )$ Illustration: Simplify: $left [left ( frac{-5}{11} right )^{1} right ]^{0}$ Solution: The given expression is $left [left ( frac{-5}{11} right )^{1} right ]^{0}$ $left [left ( frac{-5}{11} right )^{1} right ]^{0}=left [ frac{-5}{11} right ]^{0}$ $left [ because a^{1}=a right ]$ $=1$ $left [ because a^{0}=1 right ]$ Illustration: Simplify: $left [left ( frac{2}{9} right ) ^{0}right ] ^{1}$ Solution: The given expression is $left [left ( frac{2}{9} right ) ^{0}right ] ^{1}$ $left [left ( frac{2}{9} right ) ^{0}right ] ^{1}=(1)^{1}$ $left [ because a^{0}=1 right ]$ $=1$ $left [ because a^{1}=a right ]$
Positive and Negative Integral Exponents Powers with Positive Exponents: Powers with positive exponents are also known as positive integral exponents. $(10)^{0}$ = 1 $(10)^{1}$ = 10 $(10)^{2}$ = 10 × 10 = 100 $(10)^{3}$ = 10 × 10 × 10 = 1000 $(10)^{4}$ = 10 × 10 × 10 × 10 = 10000 $(10)^{5}$ = 10 × 10 × 10 × 10 × 10 = 100000 and so on Powers with Negative Exponents: For any non-zero rational 'a' and a positive integer, we define $(a)^{-n}=frac{1}{a^{n}}$ i.e. , $a^{-n}$ is the reciprocal of $a^{n}$ . Powers with negative exponents are also known as negative integral exponents. $(10)^{-1}=frac{1}{10}$ $(10)^{-2}=frac{1}{10^{2}}=frac{1}{100}$ $(10)^{-3}=frac{1}{10^{3}}=frac{1}{1000}$ $(10)^{-4}=frac{1}{10^{4}}=frac{1}{10000}$ and so on
Statement Sums Involving Exponent and Power To solve statement sums involving exponents and powers. First, frame the equation using the given values in the questions and then solve using BODMAS ( Bracket, Of, Division, Multiplication, Addition and Subtraction) Illustration 1: By what number should $left (frac{-3}{2}right )^-^3$ be divided so that the quotient is $left(frac{9}{4}right)^-^2$ ? Solution: Let the number by which we should divide $left (frac{-3}{2}right )^-^3$ be x . So, we have $frac{left(frac{-3}{2}right)^-^3}{x}=left (frac{9}{4}right)^-^2$ After cross multiplication, we have $left(frac{-2}{3}right)^3 =left(frac{4}{9}right)^2;;times;;x$ $left [ because a^{-m}=frac{1}{a^{m}} right ] ;and; left [left ( frac{a}{b} right )^{-1}=frac{b}{a} right ]$ $frac{-8}{27} =frac{16}{81};;times;;x$ $x = frac{-8}{27}timesfrac{81}{16}$ $x=frac{-3}{2}$ Illustration 2: Simplify: $frac{x^-^8z^3}{x^-^4z^6}$ Solution: The given expression is $frac{x^-^8z^3}{x^-^4z^6}$ Using laws of exponent, we have $=x^{-8-(-4)}times z^{3-6}$ $left [ because frac{a^{m}}{a^{n}}=a^{m-n} right ]$ $=x^-^4z^-^3$ Illustration: Find the values of n in the following: $(5/7)^4times(5/7)^5 = (5/7)^{2n+1}$ Solution: The given expression is $(5/7)^4times(5/7)^5 = (5/7)^{2n+1}$ $Rightarrow (5/7)^{4+5} = (5/7)^{2n+1}$ $left [ because a^{m}times a^{n} =a^{m+n}right ]$ $Rightarrow (5/7)^{9} = (5/7)^{2n+1}$ When bases are the same powers are equated. Hence by comparing the exponents of both sides, we have 9 = 2n+ 1 2n = 9 - 1 2n = 8 $n = frac{8}{2}$ n = 4 Illustration: Find the value of x for which $5^{2x}$ ÷ $5^{-3}$ $=5^{5}$ Solution: The given expression is $5^{2x}$ ÷ $5^{-3}$ $=5^{5}$ . It can be written as $frac{5^{2x}}{5^{-3}}=5^{5}$ $5^{2x-(-3)}=5^{5}$ $left [ becausefrac{a^{m}}{a^{n}} =a^{m-n}right ]$ $5^{2x+3}=5^{5}$ As the base is same, So by comparing the exponents we have 2x + 3 = 5 2x = 5 - 3 2x = 2 $x=frac{2}{2}$ x = 1
Standard Form of Exponents and Power The procedure to write very small numbers in standard form is as follows Step I: Obtain the number and see whether the number is between 1 and 10 or it is less than 1. Step II: If the number is between 1 and 10, then write it as the product of the number itself and $(10)^{0}$ . Step III: If the number is less than one, then move the decimal part to the right so that there is 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 the decimal point has been moved to the right. The number so obtained is the standard form of the given number. Illustration: Write the following numbers in the standard form: (i) 0.4579 (ii) 0.000007 (iii) 0.000000564 (iv) 216000000 Solution: (i) 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. $therefore 0.4579=4.579times (10)^{-1}$ is in the standard form. (ii) To express 0.000007 in standard form, the decimal point is moved through six places to the right so that there is just one digit on the left of the decimal point. $therefore 0.000007=7times (10)^{-6}$ (iii) To express 0.000000564 in standard form, the decimal point is moved through seven places to the right so that there is just one digit on the left of the decimal point. $therefore 0.000000564=5.64times (10)^{-7}$ (iv) To express 216000000 in standard form, the decimal point is moved through eight places to the left so that there is just one digit on the left of the decimal point. $therefore 216000000=2.16times (10)^{8}$ Illustration: Express the following numbers in the usual form (i) $small 3.52times 10^{5}$ (ii) $small 3times 10^{-5}$ Solution: (i)The given expression is $small 3.52times 10^{5}$ $small 3.52times 10^{5}=frac{352}{100}times 100000$ $small =352000$ (ii) The given expression is $small 3times 10^{-5}$ $small 3times 10^{-5}=frac{3}{10^{5}}$ $small =frac{3}{100000}$ $small =0.00003$ Illustration: Express the speed of light in the air in standard form. Solution: As we know the speed of light in air = 299705 km per second =299705000 meters per second Hence the speed of light in the air in exponential form = $2.99705times 10^{8}$ OR $3times 10^{8}$ . Illustration: Express $0.0000529times (10)^{4}$ in standard form. Solution: $0.0000529times(10)^{4}=5.29times 10^{-5}times10^{4}$ $=5.29times 10^{-5+4}$ $=5.29times 10^{-1}$

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