Rd Sharma 2019 2020 for Class 8 Maths Chapter 2 - Powers

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Page No 2.18:

Question 1:

Write each of the following in exponential form:
(i) ${(\frac{3}{2})}^{- 1} \times {(\frac{3}{2})}^{- 1} \times {(\frac{3}{2})}^{- 1} \times {(\frac{3}{2})}^{- 1}$
(ii) ${(\frac{2}{5})}^{- 2} \times {(\frac{2}{5})}^{- 2} \times {(\frac{2}{5})}^{- 2}$

Answer:

$(i) {(\frac{3}{2})}^{- 1} \times {(\frac{3}{2})}^{- 1} \times {(\frac{3}{2})}^{- 1} \times {(\frac{3}{2})}^{- 1} = {(\frac{3}{2})}^{- 1 + (- 1) + (- 1) + (- 1)} \{a^{m} \times a^{n} = a^{m + n}\} = {(\frac{3}{2})}^{- 4} (ii) {(\frac{2}{5})}^{- 2} \times {(\frac{2}{5})}^{- 2} \times {(\frac{2}{5})}^{- 2} = {(\frac{2}{5})}^{- 2 + (- 2) + (- 2)} \{a^{m} \times a^{n} = a^{m + n}\} = {(\frac{2}{5})}^{- 6}$

Page No 2.18:

Question 2:

Evaluate:
(i) 5⁻²
(ii) (−3)⁻²
(iii) ${(\frac{1}{3})}^{- 4}$
(iv) ${(\frac{- 1}{2})}^{- 1}$

Answer:

$(i) 5^{- 2} = \frac{1}{5^{2}}$               ---> (a⁻ⁿ = 1/(aⁿ))
               $= \frac{1}{25}$

$(ii) (- 3)^{2} = \frac{1}{3^{2}}$               ---> (a⁻ⁿ = 1/(aⁿ))
                         $= \frac{1}{9}$

(iii) ${(\frac{1}{3})}^{- 4} = \frac{1}{{(1 / 3)}^{4}}$              ---> (a⁻ⁿ = 1/(aⁿ))
                           $= \frac{1}{1 / 81}$
                          = 81

$(iv) {(\frac{- 1}{2})}^{- 1} = (\frac{1}{- 1 / 2})$          ---> (a⁻¹ = 1/(a))
                               $= - 2$

Page No 2.18:

Question 3:

Express each of the following as a rational number in the form $\frac{p}{q} :$
(i) 6⁻¹
(ii) (−7)⁻¹
(iii) ${(\frac{1}{4})}^{- 1}$
(iv) $(- 4)^{- 1} \times {(\frac{- 3}{2})}^{- 1}$
(v) ${(\frac{3}{5})}^{- 1} \times {(\frac{5}{2})}^{- 1}$

Answer:

$(i) 6^{- 1} = \frac{1}{6}$                 ---> (a⁻¹ = 1/a)

$(ii) (- 7)^{- 1} = \frac{1}{- 7}$              ---> (a⁻¹ = 1/a)
                        = $\frac{- 1}{7}$

$(iii) {(\frac{1}{4})}^{- 1} = \frac{1}{1 / 4}$              ---> (a⁻¹ = 1/a)
                           $= 4$

(iv) $(- 4)^{- 1} \times {(\frac{- 3}{2})}^{- 1} = \frac{1}{- 4} \times \frac{1}{- 3 / 2}$            ---> (a⁻¹ = 1/a)
                                                   $= \frac{1}{- 4} \times \frac{2}{- 3}$
                                                   $= \frac{1}{6}$

$(v)left(frac{3}{5} right )^{-1}times left(frac{5}{2} right )^{-1}=frac{1}{3/5}times frac{1}{5/2}$              ---> (a⁻¹ = 1/a)
                                               $= \frac{5}{3} \times \frac{2}{5}$
                                               $= \frac{2}{3}$

Page No 2.18:

Question 4:

Simplify:
(i) ${\{4^{- 1} \times 3^{- 1}\}}^{2}$
(ii) ${\{5^{- 1} \div 6^{- 1}\}}^{3}$
(iii) ${(2^{- 1} + 3^{- 1})}^{- 1}$
(iv) ${\{3^{- 1} \times 4^{- 1}\}}^{- 1} \times 5^{- 1}$
(v) $(4^{- 1} - 5^{- 1}) \div 3^{- 1}$

Answer:

$(i)left(4^{-1}times3^{-1} right )^2=left(frac{1}{4}times frac{1}{3}right )^2$            ---> (a⁻¹ = 1/a)
                                   $= {(\frac{1}{12})}^{2}$
                                   $= \frac{(1)^{2}}{(12)^{2}}$                          --->((a/b)ⁿ = (aⁿ)/(bⁿ))
                                   $= \frac{1}{144}$

$(ii)left(5^{-1}div6^{-1} right )^3=left(frac{1}{5}divfrac{1}{6} right)^3$          ---> (a⁻¹ = 1/a)
                                     $= {(\frac{6}{5})}^{3}$
                                     $= \frac{216}{125}$                          --->((a/b)ⁿ = (aⁿ)/(bⁿ))

$(i i i) {(2^{- 1} + 3^{- 1})}^{- 1} = {(\frac{1}{2} + \frac{1}{3})}^{- 1} {---> (a}^{- 1} = 1/a) = {(\frac{5}{6})}^{- 1} = \frac{6}{5} {---> (a}^{- 1} = 1/a)$

$(iv)left(3^{-1}times4^{-1} right )^{-1}times5^{-1}=left(frac{1}{3}timesfrac{1}{4} right )^{-1}timesfrac{1}{5}$          ---> (a⁻¹ = 1/a)
                                                     $= {(\frac{1}{12})}^{- 1} \times \frac{1}{5}$
                                                     $= 12 \times \frac{1}{5}$                               ---> (a⁻¹ = 1/a)
                                                     $= \frac{12}{5}$

$(v)left(4^{-1}-5^{-1} right )div 3^{-1}=left(frac{1}{4}-frac{1}{5} right )div frac{1}{3}$          ---> (a⁻¹ = 1/a)
                                               $=left(frac{5-4}{20} right )times 3$
                                               $= \frac{1}{20} \times 3$

                                               $= \frac{3}{20}$

Page No 2.18:

Question 5:

Express each of the following rational numbers with a negative exponent:
(i) ${(\frac{1}{4})}^{3}$
(ii) $3^{5}$
(iii) ${(\frac{3}{5})}^{4}$
(iv) ${\{{(\frac{3}{2})}^{4}\}}^{- 3}$
(v) ${\{{(\frac{7}{3})}^{4}\}}^{- 3}$

Answer:

$(i) . {(\frac{1}{4})}^{3} = {(\frac{4}{1})}^{- 3} [∵ a^{- n} = \frac{1}{a^{n}}] (ii) . {(3)}^{5} = {(\frac{1}{3})}^{- 5} [∵ a^{- n} = \frac{1}{a^{n}}] (iii) . {(\frac{3}{5})}^{4} = {(\frac{5}{3})}^{- 4} [∵ a^{- n} = \frac{1}{a^{n}}] (iv) . {\{{(\frac{3}{2})}^{4}\}}^{- 3} = {(\frac{3}{2})}^{- 12} [∵ {(a^{m})}^{n} = a^{mn}] (v) . {\{{(\frac{7}{3})}^{4}\}}^{- 3} = {(\frac{7}{3})}^{- 12} [∵ {(a^{m})}^{n} = a^{mn}]$

Page No 2.19:

Question 6:

Express each of the following rational numbers with a positive exponent:
(i) ${(\frac{3}{4})}^{- 2}$
(ii) ${(\frac{5}{4})}^{- 3}$
(iii) $4^{3} \times 4^{- 9}$
(iv) ${\{{(\frac{4}{3})}^{- 3}\}}^{- 4}$
(v) ${\{{(\frac{3}{2})}^{4}\}}^{- 2}$

Answer:

$(i) {(\frac{3}{4})}^{- 2} = {(\frac{4}{3})}^{2}$           ---> (a⁻¹ = 1/a)

$(i i) {(\frac{5}{4})}^{- 3} = {(\frac{4}{5})}^{3}$                   ---> (a⁻¹ = 1/a)

$(i i i) 4^{3} \times 4^{- 9} = 4^{(3 - 9)} = 4^{- 6} = {(\frac{1}{4})}^{6}$          ---> (a^m x aⁿ = a^m+n)

$(i v) {\{{(\frac{4}{3})}^{- 3}\}}^{- 4} = {(\frac{4}{3})}^{- 4 \times - 3} = {(\frac{4}{3})}^{12}$    ---> ((a^m)ⁿ = a^mn)

$(v) {\{{(\frac{3}{2})}^{4}\}}^{- 2} = {(\frac{3}{2})}^{4 \times - 2} = {(\frac{3}{2})}^{- 8} = {(\frac{2}{3})}^{8}$ ---> ((a^m)ⁿ = a^mn)

Page No 2.19:

Question 7:

Simplify:
(i) $\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3}$
(ii) $(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}$
(iii) ${\{{(\frac{1}{2})}^{- 1} \times (- 4)^{- 1}\}}^{- 1}$
(iv) ${[{\{{(\frac{- 1}{4})}^{2}\}}^{- 2}]}^{- 1}$
(v) ${\{{(\frac{2}{3})}^{2}\}}^{3} \times {(\frac{1}{3})}^{- 4} \times 3^{- 1} \times 6^{- 1}$

Answer:

$(i)left(left(frac{1}{3} right )^{-3}- left(frac{1}{2} right )^{-3}right )div left(frac{1}{4} right )^{-3}=left(frac{1}{(1/3)^3}- frac{1}{left (1/2 right )^3}right )divfrac{1}{(1/4)^{3}}$             ---> (a⁻ⁿ = 1/(aⁿ))
                                                                           $=left(frac{1}{(1/27)}- frac{1}{left (1/8 right )}right )divfrac{1}{(1/64)}$
                                                                           $=left(frac{27}{1}- frac{8}{1}right )div64$
                                                                           $=left(19right )timesfrac{1}{64}$
                                                                           $=frac{19}{64}$

$(ii)left(3^2-2^2right )times left(frac{2}{3} right )^{-3}=left(9-4 right )timesfrac{1}{(2/3)^3}$            ---> (a⁻ⁿ = 1/(aⁿ))
                                                   $=5timesfrac{1}{8/27}$
                                                   $=5timesfrac{27}{8}$
                                                   $=frac{135}{8}$

$(iii)left(left(frac{1}{2}right )^{-1}times left(-4 right )^{-1}right)^{-1}=left (left(frac{1}{1/2}right )times left(frac{1}{-4} right ) right )^{-1}$            ---> (a⁻¹ = 1/a)
                                                             $=left (2times left (frac{1}{-4} right ) right )^{-1}$
                                                             $=left (frac{1}{-2} right )^{-1}$
                                                             $=frac{1}{1/(-2)}$                                                ---> (a⁻¹ = 1/a)
                                                            =-2

$(iv)left(left(left(frac{-1}{4} right )^2 right )^{-2} right )^{-1}=left (left (frac{(-1)^2}{4^2} right )^{-2} right )^{-1}$           --> ((a/b)ⁿ = aⁿ/(bⁿ))
                                                    $=left (left (frac{1}{16} right )^{-2} right )^{-1}$                   ---> (a⁻ⁿ = 1/(aⁿ))
                                                    $=left (left (frac{1}{(1/16)^2} right ) right )^{-1}$
                                                    $=left (frac{1}{(1/256)} right )^{-1}$
                                                    $=256^{-1}$                                      ---> (a⁻¹ = 1/a)
                                                    $=frac{1}{256}$

$(v)left(left(frac{2}{3} right )^2 right )^3timesleft(frac{1}{3} right )^{-4}times3^{-1}times6^{-1} =left (frac{2^2}{3^2} right )^3timesfrac{1}{(1/3)^4}timesfrac{1}{3}timesfrac{1}{6}$    ---> ((a/b)ⁿ = aⁿ/(bⁿ)) and (a⁻ⁿ = 1/(aⁿ))
                                                                               $=left (frac{4}{9} right )^3timesfrac{1}{(1/81)}timesfrac{1}{3}timesfrac{1}{6}$
                                                                               $=frac{4^3}{9^3}times81timesfrac{1}{18}$                              ---> ((a/b)ⁿ = aⁿ/(bⁿ))
                                                                  $=frac{64}{729}times81timesfrac{1}{18}$
                                                                               $=frac{64}{9}timesfrac{1}{18}$
                                                                               $=64timesfrac{1}{162}$
                                                                               $=frac{64}{162}$
                                                                               $=frac{32}{81}$

Page No 2.19:

Question 8:

By what number should 5⁻¹ be multiplied so that the product may be equal to (−7)⁻¹?

Answer:

Expressing in fraction form, we get:
5⁻¹ = 1/5 (using the property a⁻¹ = 1/a)
and
(−7)⁻¹ = −1/7 (using the property a⁻¹ = 1/a).
We have to find a number x such that
$frac{1}{5}x=frac{-1}{7}$
Multiplying both sides by 5, we get:
$x=-frac{5}{7}$
Hence, 5⁻¹should be multiplied by −5/7 to obtain (−7)⁻¹.

Page No 2.19:

Question 9:

By what number should ${(\frac{1}{2})}^{- 1}$ be multiplied so that the product may be equal to ${(\frac{- 4}{7})}^{- 1} ?$

Answer:

Expressing in fractional form, we get:
(1/2)⁻¹ = 2, ---> (a⁻¹ = 1/a)
and
(−4/7)⁻¹ = −7/4 ---> (a⁻¹ = 1/a)
We have to find a number x such that
$2x=-frac{7}{4}$
Dividing both sides by 2, we get:
$x=-frac{7}{8}$
Hence, (1/2)⁻¹should be multiplied by −7/8 to obtain (−4/7)⁻¹.

Page No 2.19:

Question 10:

By what number should (−15)⁻¹ be divided so that the quotient may be equal to (−5)⁻¹?

Answer:

Expressing in fractional form, we get:
(−15)⁻¹ = −1/15,      ---> (a⁻¹ = 1/a)
and
(−5)⁻¹ = −1/5           ---> (a⁻¹ = 1/a)
We have to find a number x such that
$-frac{1}{15}div x=-frac{1}{5}$
Solving this equation, we get:
$-frac{1}{15}times frac{1}{x}=-frac{1}{5}$
           $-frac{1}{15}=-frac{x}{5}$
           $frac{-5}{-15}=x$
$therefore x=frac{1}{3}$
Hence, (−15)⁻¹should be divided by 1/3 to obtain (−5)⁻¹.

Page No 2.19:

Question 11:

By what number should ${(\frac{5}{3})}^{- 2}$ be multiplied so that the product may be ${(\frac{7}{3})}^{- 1} ?$

Answer:

Expressing as a positive exponent, we have:
$left (frac{5}{3} right )^{-2}=frac{1}{(5/3)^2}$         ---> (a⁻¹ = 1/a)
                 $=frac{1}{25/9}$             ---> ((a/b)ⁿ = (aⁿ)/(bⁿ))
                 $=frac{9}{25}$
and
(7/3) −¹ = 3/7.                ---> (a⁻¹ = 1/a)
We have to find a number x such that
$frac{9}{25}times x=frac{3}{7}$
Multiplying both sides by 25/9, we get:
$x=frac{3}{7}timesfrac{25}{9}=frac{1}{7}timesfrac{25}{3}=frac{25}{21}$
Hence, (5/3)⁻² should be multiplied by 25/21 to obtain (7/3)⁻¹.

Page No 2.19:

Question 12:

Find x, if
(i) ${(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x}$
(ii) ${(\frac{- 1}{2})}^{- 19} \times {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1}$
(iii) ${(\frac{3}{2})}^{- 3} \times {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1}$
(iv) ${(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x}$
(v) ${(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5}$
(vi) ${(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}$

Answer:

(i) We have:

$left(frac{1}{4} right )^{-4}timesleft(frac{1}{4} right )^{-8}&=&left(frac{1}{4} right )^{-4x}$
                     $left(frac{1}{4} right )^{-12}right )&=&left(frac{1}{4} right )^{-4x}$ $(a^{m} \times a^{n} = a^{m + n})$


x = 3

(ii) We have:

$left(frac{-1}{2} right )^{-19}timesleft(frac{-1}{2} right )^{8}=left(frac{-1}{2} right )^{-2x+1}$
                       $left(frac{-1}{2} right )^{-11}=left(frac{-1}{2} right )^{-2x+1}$ $(a^{m} \times a^{n} = a^{m + n})$



x = 6

(iii) We have:

$left(frac{3}{2} right )^{-3}timesleft(frac{3}{2} right )^{5}=left(frac{3}{2} right )^{2x+1}$
                       $left(frac{3}{2} right )^{2}=left(frac{3}{2} right )^{2x+1}$


                                 $frac{1}{2}=x$
x = 1/2

(iv) We have:

$left(frac{2}{5} right )^{-3}timesleft(frac{2}{5} right )^{15}=left(frac{2}{5} right )^{2+3x}$
                     $left(frac{2}{5} right )^{12}=left(frac{2}{5} right )^{2+3x}$


                              $frac{10}{3}=x$
x = 10/3

(v) We have:

$left(frac{5}{4} right )^{-x}divleft(frac{5}{4} right )^{-4}=left(frac{5}{4} right )^{5}$
                 $left(frac{5}{4} right )^{-x+4}=left(frac{5}{4} right )^{5}$



x = −1

(vi) We have:

$left(frac{8}{3} right )^{2x+1}timesleft(frac{8}{3} right )^{5}=left(frac{8}{3} right )^{x+2}$
                   $left(frac{8}{3} right )^{2x+6}=left(frac{8}{3} right )^{x+2}$


x = −4

Page No 2.19:

Question 13:

(i) If $x = {(\frac{3}{2})}^{2} \times {(\frac{2}{3})}^{- 4}$ , find the value of x⁻².
(ii) If $x = {(\frac{4}{5})}^{- 2} \div {(\frac{1}{4})}^{2}$ , find the value of x⁻¹.

Answer:

(i) First, we have to find x.

$x = {(\frac{3}{2})}^{2} \times {(\frac{2}{3})}^{- 4} = {(\frac{3}{2})}^{2} \times {(\frac{3}{2})}^{4} = {(\frac{3}{2})}^{6}$           --->(a⁻¹ = 1/a)


Hence, x⁻² is:

$x^{- 2} = {({(\frac{3}{2})}^{6})}^{- 2} = {(\frac{3}{2})}^{- 12^{}} = {(\frac{2}{3})}^{12}$            --->(a⁻¹ = 1/a)

(ii) First, we have to find x.
$x=left(frac{4}{5} right )^{-2} div left(frac{1}{4}right)^{2}$     ---> ((a/b)ⁿ = (aⁿ)/(bⁿ))
     $=left(frac{4^{-2}}{5^{-2}} right ) times4^2$
     $=frac{4^{0}}{5^{-2}}$
     $=frac{1}{5^{-2}}$                             ---> (a⁰ = 1)
Hence, the value of x⁻¹ is:
$x^{-1}=left (frac{1}{5^{-2}} right )^{-1}$
         $=left (5^2 right )^{-1}$           --->(a⁻¹ = 1/a)
         $=frac{1}{5^{2}}$                   --->(a⁻¹ = 1/a)

Page No 2.19:

Question 14:

Find the value of x for which 5^2x ÷ 5⁻³ = 5⁵.

Answer:

We have:
$5^{2x}div 5^{-3}=5^5$
         $5^{2x+3}=5^5$           ---> $a^mdiv a^n=a^{m-n}$




Hence, x is 1.

Page No 2.22:

Question 1:

Express the following numbers in standard form:
(i) 6020000000000000
(ii) 0.00000000000943
(iii) 0.00000000085
(iv) 846 × 10⁷
(v) 3759 × 10⁻⁴
(vi) 0.00072984
(vii) 0.000437 × 10⁴
(viii) 4 ÷ 100000

Answer:

To express a number in the standard form, move the decimal point such that there is only one digit to the left of the decimal point.
(i) 6020000000000000 = 6.02 x 10¹⁵      (The decimal point is moved 15 places to the left.)
(ii) 0.0000000000943 = 9.43 x 10⁻¹²     (The decimal point is moved 12 places to the right.)
(iii) 0.00000000085 = 8.5 x 10⁻¹⁰     (The decimal point is moved 10 places to the right.)
(iv) 846 x 10⁷ = 8.46 x 10² x 10⁷ = 8.46 x 10⁹     (The decimal point is moved two places to the left.)
(v) 3759 x 10⁻⁴ = 3.759 x 10³ x 10⁻⁴ = 3.759 x 10⁻¹     (The decimal point is moved three places to the left.)
(vi) 0.00072984 = 7.984 x 10⁻⁴     (The decimal point is moved four places to the right.)
(vii) 0.000437 x 10⁴ = 4.37 x 10⁻⁴ x 10⁴ = 4.37 x 10⁰ = 4.37     (The decimal point is moved four places to the right.)
(viii) 4/100000 = 4 x 100000⁻¹ = 4 x 10⁻⁵     (Just count the number of zeros in 1,00,000 to determine the exponent of 10.)

Page No 2.22:

Question 2:

Write the following numbers in the usual form:
(i) 4.83 × 10⁷
(ii) 3.02 × 10⁻⁶
(iii) 4.5 × 10⁴
(iv) 3 × 10⁻⁸
(v) 1.0001 × 10⁹
(vi) 5.8 × 10²
(vii) 3.61492 × 10⁶
(viii) 3.25 × 10⁻⁷

Answer:

(i) 4.83 x 10⁷ = 4.83 x 1,00,00,000 = 4,83,00,000
(ii) 3.02 x 10⁻⁶ = 3.02/10⁶ = 3.02/10,00,000 = 0.00000302
(iii) 4.5 x 10⁴ = 4.5 x 10,000 = 45,000
(iv) 3 x 10⁻⁸ = 3/10⁸ = 3/10,00,00,000 = 0.00000003
(v) 1.0001 x 10⁹ = 1.0001 x 1,00,00,00,000 = 1,00,01,00,000
(vi) 5.8 x 10² = 5.8 x 100 = 580
(vii) 3.61492 x 10⁶ = 3.61492 x 10,00,000 = 3614920
(viii) 3.25 x 10⁻⁷ = 3.25/10⁷ = 3.25/1,00,00,000 = 0.000000325

Page No 2.22:

Question 1:

Square of $(\frac{- 2}{3})$ is
(a) $- \frac{2}{3}$
(b) $\frac{2}{3}$
(c) $- \frac{4}{9}$
(d) $\frac{4}{9}$

Answer:

(d) 4/9
To square a number is to raise it to the power of 2. Hence, the square of (−2/3) is
$\frac{(- 2)^{2}}{3^{2}} = \frac{4}{9}$ ---> ( (a/b)ⁿ = (aⁿ)/(bⁿ) )

Page No 2.22:

Question 2:

Cube of $\frac{- 1}{2}$ is
(a) $\frac{1}{8}$
(b) $\frac{1}{16}$
(c) $- \frac{1}{8}$
(d) $\frac{- 1}{16}$

Answer:

(c) -1/8
The cube of a number is the number raised to the power of 3. Hence the cube of −1/2 is

$\frac{(- 1)^{3}}{2^{3}}$ ---> ( (a/b)ⁿ = (aⁿ)/(bⁿ)
$= \frac{- 1}{8}$

Page No 2.23:

Question 3:

Which of the following is not equal to ${(\frac{- 3}{5})}^{4} ?$
(a) $\frac{(- 3)^{4}}{5^{4}}$
(b) $\frac{3^{4}}{(- 5)^{4}}$
(c) $- \frac{3^{4}}{5^{4}}$
(d) $\frac{- 3}{5} \times \frac{- 3}{5} \times \frac{- 3}{5} \times \frac{- 3}{5}$

Answer:

(c) −(3⁴/5⁴)

${(\frac{- 3}{5})}^{4} = \frac{(- 3)^{4}}{5^{4}} = \frac{3^{4}}{(- 5)^{4}} = \frac{- 3}{5} \times \frac{- 3}{5} \times \frac{- 3}{5} \times \frac{- 3}{5}$
$It is not equal to - \frac{3^{4}}{5^{4}}$ .

Page No 2.23:

Question 4:

Which of the following is not reciprocal of ${(\frac{2}{3})}^{4} ?$
(a) ${(\frac{3}{2})}^{4}$
(b) ${(\frac{2}{3})}^{- 4}$
(c) ${(\frac{3}{2})}^{- 4}$
(d) $\frac{3^{4}}{2^{4}}$

Answer:

(c) (3/2)⁻⁴
The reciprocal of $left (frac{2}{3} right )^4$ is $left (frac{3}{2} right )^4$ .
Therefore, option (a) is the correct answer.
Option (b) is just re-expressing the number with a negative exponent.
Option (d) is obtained by working out the exponent.
Hence,option (c) is not the reciprocal of $left (frac{2}{3} right )^4$ .

Page No 2.23:

Question 5:

Which of the following numbers is not equal to $\frac{- 8}{27} ?$
(a) ${(\frac{2}{3})}^{- 3}$
(b) $- {(\frac{2}{3})}^{3}$
(c) ${(- \frac{2}{3})}^{3}$
(d) $(\frac{- 2}{3}) \times (\frac{- 2}{3}) \times (\frac{- 2}{3})$

Answer:

(a) (2/3)^-3

We can write $frac{-8}{27}$ as $frac{-2times(-2)times(-2)}{3times3times3}$ . It can be written in the forms given below.

$frac{-2times(-2)times(-2)}{3times3times3}=-frac{2times2times2}{3times3times3}$             ---> work out the minuses
                               $=-frac{2}{3}timesfrac{2}{3}timesfrac{2}{3}$
                               $=-left (frac{2}{3} right )^3$
Hence, option (b) is equal to $frac{-8}{27}$ .

We can also write:
$frac{-2times(-2)times(-2)}{3times3times3}=left (-frac{2}{3} right )timesleft (-frac{2}{3} right )timesleft (-frac{2}{3} right )$
                               $=left (-frac{2}{3} right )^3$
Hence, option (c) is also equal to $frac{-8}{27}$ .

We can also write:
$frac{-2times(-2)times(-2)}{3times3times3}=left (-frac{2}{3} right )timesleft (-frac{2}{3} right )timesleft (-frac{2}{3} right )$
Hence, option (d) is also equal to $- \frac{8}{27}$ .

This leaves out option (a) as the one not equal to $- \frac{8}{27}$ .

Page No 2.23:

Question 6:

${(\frac{2}{3})}^{- 5}$ is equal to
(a) ${(\frac{- 2}{3})}^{5}$
(b) ${(\frac{3}{2})}^{5}$
(c) $\frac{2 x - 5}{3}$
(d) $\frac{2}{3 \times 5}$

Answer:

(b) ${(\frac{3}{2})}^{5}$

Rearrange (2/3)⁻⁵ to get a positive exponent.
${(\frac{2}{3})}^{- 5} = \frac{1}{{(\frac{2}{3})}^{5}} (a^{- n} = \frac{1}{a^{n}}) = \frac{1}{\frac{2^{5}}{3^{5}}} \{{(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}\} = \frac{3^{5}}{2^{5}} = {(\frac{3}{2})}^{5}$

Page No 2.23:

Question 7:

${(\frac{- 1}{2})}^{5} \times {(\frac{- 1}{2})}^{3}$ is equal to
(a) ${(\frac{- 1}{2})}^{8}$
(b) $- {(\frac{1}{2})}^{8}$
(c) ${(\frac{1}{4})}^{8}$
(d) ${(- \frac{1}{2})}^{15}$

Answer:

(a) (−1/2)⁸

We have:

$left (frac{-1}{2} right )^5timesleft (frac{-1}{2} right )^3=left (frac{-1}{2} right )^{5+3}$
$=left (frac{-1}{2} right )^8$

Page No 2.23:

Question 8:

${(\frac{- 1}{5})}^{3} \div {(\frac{- 1}{5})}^{8}$ is equal to
(a) ${(- \frac{1}{5})}^{5}$
(b) ${(- \frac{1}{5})}^{11}$
(c) $(- 5)^{5}$
(d) ${(\frac{1}{5})}^{5}$

Answer:

(c) (−5)⁵

We have:

$left(frac{-1}{5} right )^3 div left(frac{-1}{5} right )^8 =left(frac{-1}{5} right )^{3-8}$
                                         $=left(frac{-1}{5} right )^{-5}$
                                         $=frac{1}{(-1/5)^5}$
                                         $=frac{1}{((-1)^5/5^5)}$
                                         $=frac{5^5}{(-1)^5}$
                                         $=left (frac{5}{-1} right )^5$

Page No 2.23:

Question 9:

${(\frac{- 2}{5})}^{7} \div {(\frac{- 2}{5})}^{5}$ is equal to
(a) $\frac{4}{25}$
(b) $\frac{- 4}{25}$
(c) ${(\frac{- 2}{5})}^{12}$
(d) $\frac{25}{4}$

Answer:

(a) 4/25

We have:

$left(frac{-2}{5} right )^7 div left(frac{-2}{5} right )^5 =left(frac{-2}{5} right )^{7-5}$
                                         $=left (frac{-2}{5} right )^{2}$
                                         $=frac{(-2)^2}{5^2}$
                                         $=frac{4}{25}$

Page No 2.23:

Question 10:

${\{{(\frac{1}{3})}^{2}\}}^{4}$ is equal to
(a) ${(\frac{1}{3})}^{6}$
(b) ${(\frac{1}{3})}^{8}$
(c) ${(\frac{1}{3})}^{24}$
(d) ${(\frac{1}{3})}^{16}$

Answer:

(b) (1/3)⁸

We have:

$left(left(frac{1}{3} right )^2 right )^4=left(frac{1}{3} right )^{2times4}$ ---> ( (a^m)ⁿ = a^mxn)
$=left(frac{1}{3} right )^{8}$

Page No 2.24:

Question 11:

${(\frac{1}{5})}^{0}$ is equal to
(a) 0
(b) $\frac{1}{5}$
(c) 1
(d) 5

Answer:

Page No 2.24:

Question 12:

${(\frac{- 3}{2})}^{- 1}$ is equal to
(a) $\frac{2}{3}$
(b) $- \frac{2}{3}$
(c) $\frac{3}{2}$
(d) none of these

Answer:

$(b)-frac{2}{3}$
We have:

$left (frac{-3}{2} right )^{-1}=frac{1}{(-3)/2}$           --> (a⁻¹ = 1/a)
                     $=frac{2}{-3}$

Page No 2.24:

Question 13:

${(\frac{2}{3})}^{- 5} \times {(\frac{5}{7})}^{- 5}$ is equal to
(a) ${(\frac{2}{3} \times \frac{5}{7})}^{- 10}$
(b) ${(\frac{2}{3} \times \frac{5}{7})}^{- 5}$
(c) ${(\frac{2}{3} \times \frac{5}{7})}^{25}$
(d) ${(\frac{2}{3} \times \frac{5}{7})}^{- 25}$

Answer:

$(b)left(frac{2}{3}times frac{5}{7} right ) ^{-5}$

We have:

$left(frac{2}{3} right ) ^{-5}times left(frac{5}{7} right ) ^{-5} =left(frac{2}{3}times frac{5}{7} right ) ^{-5}$ ---> ((a x b)ⁿ = aⁿ x bⁿ)

Page No 2.24:

Question 14:

${(\frac{3}{4})}^{5} \div {(\frac{5}{3})}^{5}$ is equal to

Answer:

(a) ${(\frac{3}{4} \div \frac{5}{3})}^{5}$

We have:

$left(frac{3}{4}right )^5 div left(frac{5}{3}right )^5 =left(frac{3}{4}divfrac{5}{3} right )^5$ ---> $a^{n} \div b^{n} = {(a \div b)}^{n^{}}$

Page No 2.24:

Question 15:

For any two non-zero rational numbers a and b, a⁴ ÷ b⁴ is equal to
(a) (a ÷ b)¹
(b) (a ÷ b)⁰
(c) (a ÷ b)⁴
(d) (a ÷ b)⁸

Answer:

This is one of the basic exponential formulae, i.e. .

Page No 2.24:

Question 16:

For any two rational numbers a and b, a⁵ × b⁵ is equal to
(a) (a × b)⁰
(b) (a × b)¹⁰
(c) (a × b)⁵
(d) (a × b)²⁵

Answer:

Page No 2.24:

Question 17:

For a non-zero rational number a, a⁷ ÷ a¹² is equal to
(a) a⁵
(b) a⁻¹⁹
(c) a⁻⁵
(d) a¹⁹

Answer:

Page No 2.24:

Question 18:

For a non zero rational number a, (a³)⁻² is equal to
(a) a⁹
(b) a⁻⁶
(c) a⁻⁹
(d) a¹

Answer:

(b) a⁻⁶

We have:

$left (a^3 right )^{-2}=a^{3times(-2)}$ ---> ((a^m)ⁿ = a^{m x n})
$=a^{-6}$

Page No 2.8:

Question 1:

Express each of the following as a rational number of the form $\frac{p}{q},$ where p and q are integers and q ≠ 0.
(i) 2⁻³
(ii) (−4)⁻²
(iii) $\frac{1}{3^{- 2}}$
(iv) ${(\frac{1}{2})}^{- 5}$
(v) ${(\frac{2}{3})}^{- 2}$

Answer:

We know that $a^{- n} = \frac{1}{a^{n}}$ . Therefore,

(i)
$2^{-3} = frac{1}{2^{3}}=frac{1}{8}$

(ii)

(iii)
$frac{1}{3^{-2}} = 3^2=9$

(iv)
$left (frac{1}{2} right )^{-5} = 2^5=32$

(v)
$left (frac{2}{3} right )^{-2} = left (frac{3}{2} right )^{2}=frac{9}{4}$

Page No 2.8:

Question 2:

Fiind the value of each of the following:
(i) 3⁻¹ + 4⁻¹
(ii) (3⁰ + 4⁻¹) × 2²
(iii) (3⁻¹ + 4⁻¹ + 5⁻¹)⁰
(iv) ${\{{(\frac{1}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}\}}^{- 1}$

Answer:

(i) We know from the property of powers that for every natural number a, a⁻¹ = 1/a. Then:
$3^{-1}+4^{-1}=frac{1}{3}+frac{1}{4}$           ---> (a⁻¹ = 1/a)
                       $=frac{4+3}{12}$
                       $=frac{7}{12}$

(ii) We know from the property of powers that for every natural number a, a⁻¹ = 1/a.
Moreover, a⁰ is 1 for every natural number a not equal to 0. Then:

$(3^{0} + 4^{- 1}) \times 2^{2} = (1 + \frac{1}{4}) \times 4 [as, a^{- 1} = \frac{1}{a}; a^{0} = 1] = \frac{5}{4} \times 4 = 5$

(iii) We know from the property of powers that for every natural number a, a⁻¹ = 1/a.
Moreover, a⁰ is 1 for every natural number a not equal to 0. Then:
$(3^{- 1} + 4^{- 1} + 5^{- 1}) = 1$           ---> (Ignore the expression inside the bracket and use a⁰ = 1 immediately.)

(iv) We know from the property of powers that for every natural number a, a⁻¹ = 1/a. Then:
$left (left (frac{1}{3} right )^{-1}-left (frac{1}{4} right )^{-1} right )^{-1}=left ( 3-4 right )^{-1}$           ---> (a⁻¹ = 1/a)
                                                   $=left ( -1 right )^{-1}$
                                                  = $- 1$                       ---> (a⁻¹ = 1/a)

Page No 2.8:

Question 3:

Find the value of each of the following:
(i) ${(\frac{1}{2})}^{- 1} + {(\frac{1}{3})}^{- 1} + {(\frac{1}{4})}^{- 1}$
(ii) ${(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2}$
(iii) (2⁻¹ × 4⁻¹) ÷ 2⁻²
(iv) (5⁻¹ × 2⁻¹) ÷ 6⁻¹

Answer:

(i)
${(\frac{1}{2})}^{- 1} + {(\frac{1}{3})}^{- 1} + {(\frac{1}{4})}^{- 1} = \frac{1}{1 / 2} + \frac{1}{1 / 3} + \frac{1}{1 / 4}$           --> (a⁻¹ = 1/a)
                                                           =2+3+4
                                                           =12

(ii)
${(\frac{1}{2})}^{- 2} + (\frac{1}{3})$ ^-2+14-2=11/22+11/32+11/42        --> (a⁻ⁿ = 1/(aⁿ))
                                                         = $\frac{1}{1 / 4} + \frac{1}{1 / 9} + \frac{1}{1 / 16}$               --> ((a/b)ⁿ = (aⁿ/bⁿ))
                                                         = 4+9+16
                                                          =29

(iii)
$(2^{- 1} \times 4^{- 1}) \div 2^{- 2} = (\frac{1}{2} \times \frac{1}{4}) \div \frac{1}{2^{2}}$    --> (a⁻ⁿ = 1/(aⁿ))

                                       = $\frac{1}{8} \times 4$
                                       = 2

(iv)
$(5^{- 1} \times 2^{- 1}) \div 6^{- 1} = (\frac{1}{5} \times \frac{1}{2}) \div \frac{1}{6}$              --> (a⁻ⁿ = 1/(aⁿ))
                                       $= \frac{1}{10} \times 6$

                                       $= \frac{3}{5}$

Page No 2.8:

Question 4:

Simplify:
(i) ${(4^{- 1} \times 3^{- 1})}^{2}$
(ii) ${(5^{- 1} \div 6^{- 1})}^{3}$
(iii) ${(2^{- 1} + 3^{- 1})}^{- 1}$
(iv) ${(3^{- 1} \times 4^{- 1})}^{- 1} \times 5^{- 1}$

Answer:

(i)
$left (4^{-1}times 3^{-1} right )^2=left (frac{1}{4}times frac{1}{3} right )^2$           ---> (a⁻¹ = 1/a)
                             $=left (frac{1}{12} right )^2$
                             $=frac{1^2}{12^2}$                           ---> ((a/b)ⁿ = (aⁿ)/(bⁿ) )
                             $= \frac{1}{24}$

(ii)
$left (5^{-1}div 6^{-1} right )^3=left (frac{1}{5}div frac{1}{6} right )^3$           ---> (a⁻¹ = 1/a)
                            $=left (frac{1}{5}times6 right )^3$
                            = ${(\frac{6}{5})}^{3}$
                            = $\frac{{(6)}^{3}}{{(5)}^{3}}$                          ---> ((a/b)ⁿ = (aⁿ)/(bⁿ) )

                            = $\frac{216}{125}$

(iii)
$(2^{- 1} + 3^{- 1})^{- 1} = {(\frac{1}{2} + \frac{1}{3})}^{- 1}$           ---> (a⁻¹ = 1/a)
                             = ${(\frac{5}{6})}^{- 1}$
                               $= \frac{1}{5 / 6}$                           ---> (a⁻¹ = 1/a)
                               $= \frac{6}{5}$

(iv)
$left (3^{-1}times 4^{-1} right )^{-1}times 5^{-1}=left (frac{1}{3}times frac{1}{4} right )^{-1}times frac{1}{5}$           ---> (a⁻¹ = 1/a)
                                             $=left (frac{1}{12}right )^{-1}times frac{1}{5}$
                                             $= \frac{12}{5}$                                       ---> (a⁻¹ = 1/a)

Page No 2.8:

Question 5:

Simplify:
(i) $(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3}$
(ii) $(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}$
(iii) $[{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}] \div {(\frac{1}{4})}^{- 3}$
(iv) $(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2}$

Answer:

(i)
$(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3} = (9 + 4) \times \frac{1}{8} = \frac{13}{8}$

(ii)
$left ( 3^2-2^2 right )times left(frac{2}{3} right )^{-3}=left(9-4 right )times frac{1}{left (2/3 right )^{3}}$             ---> (a⁻¹=1/(aⁿ))
                                           $=5times frac{1}{8/27}$                               ---> ((a/b)ⁿ = (aⁿ)/(bⁿ))
                                           $= 5 \times \frac{27}{8}$
                                           $= \frac{135}{8}$

(iii)
$left( left(frac{1}{3} right )^{-3}-left(frac{1}{2} right )^{-3}right )div left( frac{1}{4}right )^{-3}=left(3^3-2^3 right )div 4^3$             --->(a^-n = 1/(aⁿ))
                                                                  = $(27 - 8) \div 64$
                                                                   $= 19 \times \frac{1}{64}$
                                                                   $= \frac{19}{64}$

(iv)
$(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2} = (4 + 9 - 16) \times \frac{9}{4}$                     ---> ((a/b)ⁿ = (aⁿ)/(bⁿ))

                                                   $= - 3 \times \frac{9}{4}$
                                                  $= \frac{- 27}{4}$

Page No 2.8:

Question 6:

By what number should 5⁻¹ be multiplied so that the product may be equal to (−7)⁻¹?

Answer:

Using the property a⁻¹ = 1/a for every natural number a, we have 5⁻¹ = 1/5 and (−7)⁻¹ = −1/7. We have to find a number x such that
$\frac{1}{5} \times x = \frac{- 1}{7}$
Multiplying both sides by 5, we get:
$x = \frac{- 5}{7}$
Hence, the required number is −5/7.

Page No 2.8:

Question 7:

By what number should ${(\frac{1}{2})}^{- 1}$ be multiplied so that the product may be equal to ${(- \frac{4}{7})}^{- 1} ?$

Answer:

Using the property a⁻¹ = 1/a for every natural number a, we have (1/2)⁻¹ = 2 and (−4/7)⁻¹ = −7/4. We have to find a number x such that
$2 x = \frac{- 7}{4}$
Dividing both sides by 2, we get:
$x = \frac{- 7}{8}$
Hence, the required number is −7/8.

Page No 2.8:

Question 8:

By what number should (−15)⁻¹ be divided so that the quotient may be equal to (−5)⁻¹?

Answer:

Using the property a⁻¹ = 1/a for every natural number a, we have (−15)⁻¹ = −1/15 and (−5)⁻¹ = −1/5. We have to find a number x such that
$\frac{\frac{- 1}{15}}{\frac{x}{1}} = \frac{- 1}{5} or \frac{- 1}{15} \times \frac{1}{x} = \frac{- 1}{5} or x = \frac{1}{3}$
Hence, (−15)⁻¹should be divided by $\frac{1}{3}$ to obtain (−5)⁻¹.

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