RD Sharma solution class 7 chapter 6 Exponents Exercise 6.2

Exercise 6.2

Page-6.28

Question 1:

Using laws of exponents, simplify and write the answer in exponential form:
(i) 2³ × 2⁴ × 2⁵
(ii) 5¹² ​÷ 5³
(iii) (7²)³
(iv) (3²)⁵ ​÷ 3⁴
(v) 3⁷ ​× 2⁷
(vi) (5²¹ ÷ 5¹³) × 5⁷

Answer 1:

 We have

(i)  2³ x 2⁴ x 2⁵ = 2^{(3 + 4 + 5)} = 2¹²             [since a^m + aⁿ + a^p = a^(m+n+p)]

(ii) 5¹² ÷ 5³ = $\frac{5^{12}}{5^3}$ = 5^{12 - 3} = 5⁹                [ since a^m ÷ aⁿ = a^m-n ]
(iii) (7²)³ = 7⁶                                                       [since (a^m)ⁿ = a^mn ]

(iv)(3²)⁵ ÷ 3⁴ = 3¹⁰ ÷ 3⁴                             [since (a^m)ⁿ = a^mn ]
                       = 3^{(10 - 4)} = 3⁶                      [since a^m ÷ aⁿ = a^m-n ]

(v) 3⁷ ​× 2⁷ = (3 x 2)⁷ = 6⁷                          [since a^m x b^m = (a x b)^m ]

 (vi) (5²¹ ÷  5¹³) x 5⁷ = 5^{(21 -13)} x 5⁷         [since a^m ÷ aⁿ = a^m-n ]
                                   = 5⁸ x 5⁷                  [since a^m x bⁿ =a^{(m +n)}]
                                   = 5⁽⁸⁺⁷⁾              
                                   = 5¹⁵

Question 2:

Simplify and express each of the following in exponential form:
(i) $\left\{\right(2^3{)}^4\times 2^8\}÷2^{12}$
(ii) (8² × 8⁴) ÷ 8³
(iii) $\left(\frac{5^7}{5^2}\right)\times 5^3$
(iv) $\frac{5^4\times x^{10}{y}^5}{5^4\times x^7{y}^4}$

Answer 2:

We have
(i)  {(2³)⁴ x 2⁸} ÷ 2¹²
    = {2¹² x 2⁸} ÷ 2¹²
    = 2^{(12 + 8)} ÷ 2¹²
    = 2²⁰ ÷ 2¹²
    = 2 ^{(20 - 12)} =  2⁸

(ii) (8² x 8⁴)  ÷ 8³
    = 8^{(2 + 4)} ÷ 8³
    = 8⁶ ÷ 8³
   = 8^(6-3) = 8³ = (2³)³ = 2⁹

(iii) $\left(\frac{5^7}{5^2}\right)$ x 5³ = 5^(7-2) x 5³
                        = 5⁵ x 5³
                        = 5^{(5 + 3 )} = 5⁸

(iv) $\frac{5^4 \times x^{10}{y}^5}{5^4\times x^7{y}^4}$ = $5^{(4-4)}\times x^{(10-7)}\times y^{(5-4)}$
                         = $5^0\times x^3\times y$               [since 5⁰ = 1]
                         = $1\times x^{3}y =x^{3}y$

Question 3:

Simplify and express each of the following in exponential form:
(i) $\left\{\right(3^2{)}^3\times 2^6\}\times 5^6$
(ii) ${\left(\frac{x}{y}\right)}^{12}\times y^{24}\times (2^3{)}^4$
(iii) ${\left(\frac{5}{2}\right)}^6\times {\left(\frac{5}{2}\right)}^2$
(iv) ${\left(\frac{2}{3}\right)}^5\times {\left(\frac{3}{5}\right)}^5$

Answer 3:

We have
(i)   {(3²)³ x 2⁶} x 5⁶
    = {3⁶ x 2⁶} x 5⁶             [since (a^m)ⁿ = a^mn]
    = 6⁶ x 5⁶                        [since a^m x b^m = (a x b)^m ]
    = 30⁶

(ii)  
${\left(\frac{x}{y}\right)}^{12}\times y^{24}\times (2^3{)}^4$
$=\frac{x^{12}}{y^{12}}\times y^{24}\times 2^{12}$
$=x^{12}\times \frac{y^{24}}{y^{12}}\times 2^{12}$
$=x^{12}\times y^{24-12}\times 2^{12}$
$=x^{12}\times y^{12}\times 2^{12}$
$=(2xy{)}^{12} [\mathrm{since} a^m\times b^m\times c^m=(a\times b\times c{)}^m]$
(iii)
${\left(\frac{5}{2}\right)}^6\times {\left(\frac{5}{2}\right)}^2$
$={\left(\frac{5}{2}\right)}^8 [\mathrm{since} a^m\times a^n=a^{m+n}]$

(iv)
${\left(\frac{2}{3}\right)}^5\times {\left(\frac{3}{5}\right)}^5$
$={\left(\frac{2}{3}\times \frac{3}{5}\right)}^5 [\mathrm{since} a^m\times b^m=(a\times b{)}^m]$
$={\left(\frac{2}{5}\right)}^5$

Question 4:

Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.

Answer 4:

We have
9 x 9 x 9 x 9 x 9  = (9)⁵ =(3²)⁵ = 3¹⁰

Question 5:

Simplify and write each of the following in exponential form:
(i) (25)³ ÷ 5³
(ii) (81)⁵ ÷ (3²)⁵
(iii) $\frac{9^8\times (x^2{)}^5}{(27{)}^4\times (x^3{)}^2}$
(iv) $\frac{3^2\times 7^8\times {13}^6}{{21}^2\times {91}^3}$

Answer 5:

We have
(i) (25)³ ÷ 5³
= (5²)³÷ 5³
= 5⁶ ÷ 5³
= $\frac{5^6}{5^3}=5^{6-3}=5^3$

(ii) (81)⁵ ÷ (3²)⁵
= (3⁴)⁵ ÷ (3²)⁵
= (3)²⁰ ÷ (3)¹⁰
= $\frac{3^{20}}{3^{10}}=3^{20-10}=3^{10}$

(iii)

$\frac{9^8\times (x^2{)}^5}{(27{)}^4\times (x^3{)}^2}$
$=\frac{(3^2{)}^8\times (x^2{)}^5}{(3^3{)}^4\times (x^3{)}^2}$
$=\frac{3^{16}\times (x{)}^{10}}{3^{12}\times (x{)}^6}$
$=3$^16-12× (x)^10-6 = 3⁴× x⁴= (3x)⁴

(iv)

$\frac{3^2\times 7^8\times {13}^6}{{21}^2\times {91}^3}$
$=\frac{3^2\times 7^2\times 7^6\times {13}^6}{{21}^2\times (13\times 7{)}^3}$
$=\frac{(21{)}^2\times 7^6\times {13}^6}{{21}^2\times {13}^3\times 7^3}$
$=\frac{7^6\times {13}^6}{{13}^3\times 7^3}$
$=\frac{{91}^6}{{91}^3}={91}^{6-3}={91}^3$

Question 6:

Simplify:
(i) $(3^5{)}^{11}\times (3^{15}{)}^4-(3^5{)}^{18}\times (3^5{)}^5$
(ii) $\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}$
(iii) $\frac{10\times 5^{n+1}+25\times 5^n}{3\times 5^{n+2}+10\times 5^{n+1}}$
(iv) $\frac{(16{)}^7\times (25{)}^5\times (81{)}^3}{(15{)}^7\times (24{)}^5 \times (80{)}^3}$

Answer 6:

We have
(i) (3⁵)¹¹× (3¹⁵)⁴- (3⁵)¹⁸×(3⁵)⁵
   = 3⁵⁵ x 3⁶⁰ - 3⁹⁰ x 3²⁵
   = 3^{(55 + 60)} - 3^{(90 + 25)}
   = 3¹¹⁵ - 3¹¹⁵
   = 0

(ii) $\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}$
    =  $\frac{2^{4 }\times 2^{n+1}-2^2\times 2^n}{2^{4 }\times 2^{n+2}-2^{n+1}\times 2^2}$
   $=\frac{2^2\times (2^{n+3}-2^n)}{2^2\times (2^{n+4}-2^{n+1})}$
$=\frac{2^n\times 2^3-2^n}{2^n\times 2^4-2^n\times 2}$
$=\frac{2^n(2^3-1)}{2^n(2^4-2)}=\frac{8-1}{16-2}=\frac{7}{14}=\frac{1}{2}$

 (iii)

$\frac{10\times 5^{n+1}+25\times 5^n}{3\times 5^{n+2}+10\times 5^{n+1}}$
$=\frac{10\times 5^{n+1}+(5{)}^2\times 5^n}{3\times 5^{n+2}+2\times 5\times 5^{n+1}}$
$=\frac{10\times 5^{n+1}+5\times 5^{n+1}}{3\times 5^{n+2}+2\times 5\times 5^{n+1}}$
$=\frac{5^{n+1}(10+5)}{3\times 5\times 5^{n+1}+10\times 5^{n+1}}$
$=\frac{5^{n+1}\left(15\right)}{5^{n+1}(15+10)}=\frac{5^{n+1}\times 15}{5^{n+1}\times 25}=\frac{15}{25}=\frac{3}{5}$


(iv)
$\frac{(16{)}^7\times (25{)}^5\times (81{)}^3}{(15{)}^7\times (24{)}^5\times (80{)}^3}$
$=\frac{(16{)}^7\times (5^2{)}^5\times (3^4{)}^3}{(3\times 5{)}^7\times (3\times 8{)}^5\times {\left(16\times 5\right)}^3}$
$=\frac{(16{)}^7\times (5{)}^{10}\times (3{)}^{12}}{3^7\times 5^7\times 3^5\times 8^5\times {16}^3\times 5^3}$
$=\frac{(16{)}^7\times (5{)}^{10}\times (3{)}^{12}}{3^7\times 3^5\times 5^7\times 5^3\times 8^5\times {16}^3}$
$=\frac{(16{)}^7\times (5{)}^{10}\times (3{)}^{12}}{3^{12}\times 5^{10}\times 8^5\times {16}^3}$
$=\frac{(16{)}^7}{8^5\times {16}^3}$
$=\frac{(16{)}^{7-3}}{8^5}=\frac{(16{)}^4}{8^5}=\frac{(2\times 8{)}^4}{8^5}=\frac{2^4\times 8^4}{8^5}=\frac{2^4}{8}=\frac{16}{8}=2$

Question 7:

Find the values of n in each of the following:
(i) $5^{2n}\times 5^3=5^{11}$
(ii) $9\times 3^n=3^7$
(iii) $8\times 2^{n+2}=32$
(iv) $7^{2n+1}÷49=7^3$
(v) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{3}{2}\right)}^5={\left(\frac{3}{2}\right)}^{2n+1}$
(vi) ${\left(\frac{2}{3}\right)}^{10}+{\left\{{\left(\frac{3}{2}\right)}^2\right\}}^5={\left(\frac{2}{5}\right)}^{2n-2}$

Answer 7:

We have

(i) 5²ⁿ x 5³ = 5¹¹
= 5²ⁿ⁺³ = 5¹¹
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n =$\frac{8}{2}=4$
(ii) 9 x 3ⁿ = 3⁷
= (3)² x 3ⁿ = 3⁷
= (3)²⁺ⁿ = 3⁷
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 - 2  = 5

(iii) 8 x 2ⁿ⁺² = 32
= (2)³ x 2ⁿ⁺² = (2)⁵      [since 2³ = 8 and 2⁵ = 32]
= (2)³⁺ⁿ⁺² = (2)⁵ 
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0

(iv) 7²ⁿ⁺¹ ÷ 49 = 7³
= 7²ⁿ⁺¹ ÷ 7² = 7³  [since 49 = 7²]
$=\frac{7^{2n+1}}{7^2}=7^3$
$=7^{2n+1-2}=7^3 [\mathrm{since} \frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}]$
= 7^2n-1 =7³
On equating the coefficients, we get
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n = $\frac{4}{2}=2$
(v) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{3}{2}\right)}^5={\left(\frac{3}{2}\right)}^{2n+1}$
$={\left(\frac{3}{2}\right)}^{(4+5)}={\left(\frac{3}{2}\right)}^{(2n+1)}$
$={\left(\frac{3}{2}\right)}^9={\left(\frac{3}{2}\right)}^{2n+1}$
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 - 1
⇒ 2n = 8
⇒ n =$\frac{8}{2}=4$
 (vi) ${\left(\frac{2}{3}\right)}^{10}\times {\left\{{\left(\frac{3}{2}\right)}^2\right\}}^5={\left(\frac{2}{5}\right)}^{2n-2}$
$={\left(\frac{2}{3}\right)}^{10}\times {\left(\frac{3}{2}\right)}^{10}={\left(\frac{2}{5}\right)}^{2n-2}$
$=\frac{2^{10}\times 3^{10}}{3^{10}\times 2^{10}}={\left(\frac{2}{5}\right)}^{2n-2}$
$=1={\left(\frac{2}{5}\right)}^{2n-2}$
$={\left(\frac{2}{5}\right)}^0={\left(\frac{2}{5}\right)}^{2n-2} [\mathrm{since} {\left(\frac{2}{5}\right)}^0=1]$
On equating the coefficients, we get
⇒ 0 = 2n - 2
⇒ 2n = 2
⇒ n = $\frac{2}{2}=1$

Page-6.29

Question 8:

If $\frac{9^n\times 3^2\times 3^n-(27{)}^n}{\left(3^3\right)\times 2^3}=\frac{1}{27}$, find the value of n.

Answer 8:

We have

$\frac{9^n\times 3^2\times 3^n-(27{)}^n}{(3^3{)}^5\times 2^3}=\frac{1}{27}$
$=\frac{(3^2{)}^n\times 3^2\times 3^n-(3^3{)}^n}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{2n+2+n}-(3{)}^{3n}}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n+2}-(3{)}^{3n}}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n}\times (3{)}^2-(3{)}^{3n}}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{\left(3{)}^{3n}\right(3^2-1)}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n} \times 8}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n} \times 2^3}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{3^{3n}}{3^{15}}=\frac{1}{27}$
$=3^{3n-15}=\frac{1}{3^3}$
$=3^{3n-15}=3^{-3}$
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n =$\frac{12}{3}=4$