myhelper: RD Sharma solution class 8 chapter 2 Powers Exercise 2.2

Exercise 2.2

Page-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 1:

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

Question 2:

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

Answer 2:

$(i) 5^{- 2} = \frac{1}{5^{2}}$               ---> (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))
                               $= - 2$

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 3:

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

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 4:

$(i)left(4^{-1}times3^{-1} right )^2=left(frac{1}{4}times frac{1}{3}right )^2$            ---> (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)
                                               $=left(frac{5-4}{20} right )times 3$
                                               $= \frac{1}{20} \times 3$

                                               $= \frac{3}{20}$

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 5:

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

Page-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 6:

$(i) {(\frac{3}{4})}^{- 2} = {(\frac{4}{3})}^{2}$           ---> (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)

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 7:

$(i)left(left(frac{1}{3} right )^{-3}- left(frac{1}{2} right )^{-3}right )div left(frac{1}{4} right )^{-3}=left(frac{1}{(1/3)^3}- frac{1}{left (1/2 right )^3}right )divfrac{1}{(1/4)^{3}}$             ---> (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))
                                                                  $=frac{64}{729}times81timesfrac{1}{18}$
                                                                               $=frac{64}{9}timesfrac{1}{18}$
                                                                               $=64timesfrac{1}{162}$
                                                                               $=frac{64}{162}$
                                                                               $=frac{32}{81}$

Question 8:

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

Answer 8:

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.

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 9:

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.

Question 10:

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

Answer 10:

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.

Question 11:

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

Answer 11:

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.

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 12:

(i) We have:

$left(frac{1}{4} right )^{-4}timesleft(frac{1}{4} right )^{-8}&=&left(frac{1}{4} right )^{-4x}$
                     $left(frac{1}{4} right )^{-12}right )&=&left(frac{1}{4} right )^{-4x}$ $(a^{m} \times a^{n} = a^{m + n})$
                             -12=-4x
                                   3=x
x = 3

(ii) We have:

$left(frac{-1}{2} right )^{-19}timesleft(frac{-1}{2} right )^{8}=left(frac{-1}{2} right )^{-2x+1}$
                       $left(frac{-1}{2} right )^{-11}=left(frac{-1}{2} right )^{-2x+1}$ $(a^{m} \times a^{n} = a^{m + n})$
                                   -11=-2x+1
                                   -12=-2x
                                         6=x
x = 6

(iii) We have:

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

(iv) We have:

$left(frac{2}{5} right )^{-3}timesleft(frac{2}{5} right )^{15}=left(frac{2}{5} right )^{2+3x}$
                     $left(frac{2}{5} right )^{12}=left(frac{2}{5} right )^{2+3x}$
                              12=2+3x
                              10=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+4=5
                               -x=1
                                   x=-1
x = −1

(vi) We have:

$left(frac{8}{3} right )^{2x+1}timesleft(frac{8}{3} right )^{5}=left(frac{8}{3} right )^{x+2}$
                   $left(frac{8}{3} right )^{2x+6}=left(frac{8}{3} right )^{x+2}$
                         2x+6=x+2
                                   x=-4
x = −4

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 13:

(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)

Question 14:

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

Answer 14:

We have:
$5^{2x}div 5^{-3}=5^5$
         $5^{2x+3}=5^5$           ---> $a^mdiv a^n=a^{m-n}$
      2x+3=5
               2x=2
                 x=1

Hence, x is 1.