Chapter – 02 : Laws of Indices | Chapter Solution Class 9

(\sqrt[5]{8} )^{5/2}×16^3/2

= (8^1/5)^5/2 × (2⁴)^-3/2

= (8)^{1/5 × 5/2} × (2)^4×-3/2

= (2³)^1/2 × (2)^-6

= 2^{3 × 1/2} × 2^-6

= 2 ^{\frac{3}{2}-6}

= 2 ^{\frac{3-12}{2}}

= 2 ^{\frac{-9}{2}}

= \big\{(5^3)^{-2} \times (2^4)^{-\frac{3}{2}}\big\} ^{ -\frac{1}{6} }

= \big\{(5)^{3 \times -2} \times (2^4)^{-\frac{3}{2}}\big\} ^{ -\frac{1}{6}}

= \big\{5^{-6} \times 2{-6}\big\}^{-\frac{1}{6}}

= \big\{(5 \times 2 ^{-6} \big\}^{-\frac{1}{6}}

= (5 \times 2)^{-6 \times -\frac{1}{6} }

= (10)¹ = 10

4^\frac{1}{3}× [2^\frac{1}{3} × 3^\frac{1}{2}] \div 9^\frac{1}{4}

= \frac{4^\frac{1}{3}\times [2^\frac{1}{3} \times 3^\frac{1}{2}]}{9^\frac{1}{4}}

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

= \frac{2^{2 \times \frac{1}{3}} \times 2^\frac{1}{3} \times 3^{1/2} }{{3^{2 \times \frac{1}{4}} }}

= \frac{2^{\frac{2}{3}} \times 2^\frac{1}{3} \times 3^\frac{1}{2}}{3^{\frac{1}{2}}}

= \frac{2^{ \frac{2}{3} + \frac{1}{3} } \times 1}{1}

= 2^{\frac{2+1}{3} }

= 2^\frac{3}{3} = 2¹ = 2

= (\frac{8a{^3}}{27x^{-3}})^\frac{2}{3} \times (\frac{16a{^3}}{27x^{-3}})^{-\frac{2}{3}}

= (\frac{2^3a{^3}}{3^3x^{-3}})^\frac{2}{3} \times (\frac{4^3a{^3}}{3^3x^{-3}})^{-\frac{2}{3}}

= (\frac{2a}{3x^{-1}})^{3 \times \frac{2}{3}} \times (\frac{4a}{3x^{-1}})^{3 \times \frac{2}{3}}

= (\frac{2a}{3. \frac{1}{x}})^2 \times (\frac{4a}{3. \frac{1}{x}})^{-2}

= (\frac{2ax}{3})^2 \times (\frac{4ax}{3})^{-2}

= (\frac{2ax}{3})^2 \times (\frac{3}{4ax})^{-2}

= \frac{4^a2x^2}{9} \times \frac{9}{16a^2x^2}

= \frac{1}{4}

= \big\{x^{-5 \times {2\over3}}\big\}^{-3\over10}

= \big\{x^{-10\over3}\big\}^{-3\over10}

= x^{-{10\over3x}×{-3\over10}} = x

[{(2^-1)^-1}^-1]^-1

= {(2)^-1}^-1

= 2

= a^{-2\over3} . b \times b^{-2\over3} .c \times c^{-2\over3} . a

= a^{-2\over3} . a \times b^{-2\over3} .b \times c^{-2\over3} . c

= a^{-{2\over3}+1} . a \times b^{-{2\over3}+1} .b \times c^{-{2\over3}+1} . c

= a^{ \frac{1}{3}} . a \times b^{ \frac{1}{3}} .c^{ \frac{1}{3}} = (abc)^{ \frac{1}{3}}

Given: \frac{4^{m+\frac{1}{4}} \times \sqrt{2.2^m}^{\frac{1}{m}}}{2. \sqrt{2^{-m}} }

= (\frac{2^{2{(m+\frac{1}{4}})} \times \sqrt{2^{m+1}}}{2. {2^{-m/2}}})^\frac{1}{m}

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

= (\frac{2^{2{m+\frac{2}{4}}} \times {2^{ \frac{m+1}{2}}}}{{2{ \frac{2-m}{2}}}})^\frac{1}{m}

= (\frac{2^{\frac{8m+2+2m+2}{4}}}{2^{\frac{2-m}{2}}})^\frac{1}{m}

= (2^{\frac{10m+4}{4} - \frac{2-m}{2}})^\frac{1}{m}

= (2^\frac{10m+4-4+2}{4})^\frac{1}{m}

= (2^{ \frac{12}{4}})^\frac{1}{m}

= (2^{3m})^\frac{1}{m}

= 2^{3m \times \frac{1}{m} } = 2³ = 8

Given: 9^{-3} \times \frac{16^{\frac{1}{4}}}{6^{-2}} \times ( \frac{1}{27})^{- \frac{4}{3}}

= 9^{-3} \times \frac{16^{\frac{1}{4}}}{6^{-2}} \times ( \frac{1}{27})^{- \frac{4}{3}}

= (3^2)^{-3} \times \frac{(2^4)^{ \frac{1}{4}}}{(2 \times 3)^{-2}} \times (\frac{1}{3^3})^{-4/3}

= (3)^{2 \times -3} \times \frac{(2^4)^{ \frac{1}{4}}}{2 \times 3^{-2}} \times (\frac{1}{3})^{3 \times -4/3}

= 3^{-6} \times 2 \times 2^2 \times 3^2 \times (1/3)^{-4}

= 3^{-6} \times 2^{1+2} \times 3^2 \times (3^{-1})^{-4}

= 3^{-6} \times 2^{3} \times 3^2 \times 3^4

= 2^3 \times 3^{2+4-6}

= 2^3 \times 1

= 8 \times 1

= 8 Ans.

Given: (\frac{x^a}{x^b} )^{a^2+ab+b^2} \times ( \frac{x^b}{x^c} )^{b^2+bc+c^2} \times ( \frac{x^c}{x^a} )^{c^2+ca+a^2}

= \left(x^{a-b}\right)^{a^2 + a \cdot b + b^2} \times \left(x^{b-c}\right)^{b^2 + b \cdot c + c^2} \times \left(x^{c-a}\right)^{c^2 + c \cdot a + a^2}

= x^{(a - b)\left(a^2 + a \cdot b + b^2\right)} \times x^{(b - c)\left(b^2 + b \cdot c + c^2\right)} \times x^{(c - a)\left(c^2 + c \cdot a + a^2\right)}

= x^{a^3-b^3} \times x^{b^3-c^3} \times x^{c^3-a^3}

= x^{a^3-b^3+b^3-c^3+c^3-a^3}

= x⁰ = 1

(i) 5^{1 \over 2}, 10^{1 \over 4}, 6^{1 \over 3}

First, we have to find the L.C.M of denominators of Power 2, 4 and 3.

∴ L.C.M of 2, 4, and 3 = 12

∴ In ascending order, 5^{1 \over 2}, 6^{1 \over 3}, 10^{1 \over 4} Ans.

(ii) 3^{1 \over 3}, 2^{1 \over 2}, 8^{1 \over 4}

First, we have to find the L.C.M of denominators of Power 3, 2 and 4.

∴ L.C.M of 3, 2, and 4 = 12

∴ In ascending order, 2^{1 \over 2}, 3^{1 \over 3}, 8^{1 \over 4} Ans.

(iii) 2^{60}, 3^{48}, 4^{36}, 5^{24}

2^{60}=\left(2^5\right)^{12}=(32)^{12} \\

3^{48}=\left(3^4\right)^{12}=(81)^{12} \\

4^{36}=\left(4^3\right)^{12}=(64)^{12} \\

5^{24}=\left(5^2\right)^{12}=(25)^{12}

∴ In ascending order, 5^{24}, 2^{60}, 4^{36}, 3^{48}

Given: L.H.S= \left(\frac{a^q}{a^r}\right)^p \times\left(\frac{a^r}{a^p}\right)^q \times\left(\frac{a^p}{a^q}\right)^r \\

= \left(a^{q-r}\right)^p \times\left(a^{r-p}\right)^q \times\left(a^{p-q}\right)^r \\

= a^{p q-p r} \times a^{r q-p q} \times a^{p r-q r} \\

= a^{p q-p r+q r-p q+p r-q r} = a^0=1

∴ LHS = RHS

Given: \left(\frac{x^m}{x^n}\right)^{m+n} \times\left(\frac{x^n}{x^l}\right)^{n+l} \times\left(\frac{x^l}{x^m}\right)^{l+m}=1 \\

= \left(x^{m-n}\right)^{m+n} \times\left(x^{n-l}\right)^{n+1} \times\left(x^{l-m}\right)^{l+m}

= x^{(m-n)(m+n)} \times x^{(n-l)(n+l)} \times x^{(l-m)(l+m)} \\

= x^{m^2-n^2} \times x^{n^2-l^2} \times x^{l^2-m^2} \\

= x^{m^2-n^2+n^2-1^2+l^2-m^2} \\

= x⁰ = 1

L.H.S = R.H.S. (proved)

Given: \left(\frac{x^m}{x^n}\right)^{m+n+l} \times\left(\frac{x^n}{x^l}\right)^{n+l+m}\left(\frac{x^{l}}{x^m}\right)^{l+m+n} \\

= \left(x^{\mathrm{m}-\mathrm{n}}\right)^{\mathrm{m}+\mathrm{n}+l} \times\left(\mathrm{x}^{\mathrm{n}-l}\right)^{\mathrm{n}+l+\mathrm{m}} \times\left(\mathrm{x}^{l-\mathrm{m}}\right)^{1+\mathrm{m}+\mathrm{n}}

= \left(x^{(m-n)(m+n+l)} \times x^{(n-l)(n+l+m)} \times x^{(l-m)(l+m+n)}\right. \\

= \text{x}^{\text{(m²+m p+l m-m n-n²-l n )}} \times \text{x}^{\text{(n² + nl + mn - ln - l² - ml)}} \times \text{x}^{\text{(l² + lm + ln - ml - m² - mn)}}

= \mathrm{x}^{\mathrm{m}^2+l \mathrm{~m}-\mathrm{n}^2-l n} \times \mathrm{x}^{\mathrm{n}^2+\mathrm{mn}-l^2-l \mathrm{~m}} \times \mathrm{x}^{l^2+l n-\mathrm{m}^2-\mathrm{mn}} \\

= x^{m^2+l m-n^2-l n+n^2+m n-l^2-l m+l^2-l n-m^2-m n} \\

= x⁰ = 1

L.H.S = R.H.S. (proved)

Given: \left(a^{\frac{1}{x-y}}\right)^{\frac{1}{x-z}} \times\left(a^{\frac{1}{y-z}}\right)^{\frac{1}{y-x}} \times\left(a^{\frac{1}{z-x}}\right)^{\frac{1}{z-y}}=1 \\

= a^{\frac{1}{(x-y)(x-z)}-\frac{1}{(y-z)(x-y)}-\frac{1}{(x-z)(y-z)}} \\

= a^{\frac{1}{(x-y)(x-z)}-\frac{1}{(y-z)(x-y)}-\frac{1}{(x-z)(y-z)}} \\

= a^{\frac{y-z-(x-z)+x-y}{(x-y)(x-z)(y-z)}} \\

= \frac{y-z-x+z+x-y}{a^{(x-y)(x-z)(y-z)}} \\

= a⁰ = 1

L.H.S = R.H.S. (proved)

We have,

x + z = 2y or x + z = y + y

x – y = y – z …… (i)

L.H.S: a ^{y – z} b^{z – x} c^{x – y}

= a ^{y – z} b^{z – x} c^{y – z}

= (ac)^{y – z} b^{z – x}

= (b²)^{y – z} b^{z – x}   (since b² = ac)

= b^{2y – 2z} b^{z – x}

= b ^{2y – 2z + z – x}

= b ^{2y – (x + z)}

= b ^{2y – 2y} = b⁰ = 1 : RHS

LHS: \quad a^{q-r} b^{r-p} c^{p-q} \\

= \left(x y^{p-1}\right)^{q-r} \cdot\left(x y^{q-1}\right)^{r-p} \cdot\left(x y^{r-1}\right)^{p-q} \\

= x^{q-r} \cdot y^{(p-1)(q-r)} \cdot x^{r-p} \cdot y^{(q-1)(r-p)} \cdot x^{p-q} \cdot y^{(r-1)(p-c)} \\

= x^{q-r} \cdot y^{p q-p r-q+r} \cdot x^{r-p} \cdot y^{q r-p q-r+p)} \cdot x^{p-q} \cdot y^{p r-q r-p+q} \\

= x^{q-r+r-p+p-q} \cdot y^{p q-p r-q+r+q r-p q-r+p+p r-q r-p+q} \\

= x^0 \cdot y^0=1.1=1

L.H.S = R.H.S. (proved)

Let x^{\frac{1}{a}} = y^{\frac{1}{b}} = z^{\frac{1}{c}} = k (Say)

∴ x = k^a, y = k^b. z = k^c

and xyz = 1

or, k^a .k^b .k^c = 1

or, k^a+b+c = k⁰

Since base is same, power must be equal.

a + b + c = 0 (Proved)

Let a^x= b^y= c^z = k (Say)

a = k^1/x ; b = k^1/y ; c = k^1/z

Given: abc = 1

or, \quad k^{1 \over x} \cdot k^{1 \over y} \cdot k^{1 \over z} \\ = 1

\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}} = k⁰   [since k⁰ = 1]

Since base is same, power must be equal.

∴ \quad \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=0 \\

or, \quad \frac{\text{y z+z x+x y}}{\text{x y z}} = 0

∴ xy + yz + zx = 0 (Proved)

Given: 49^x = 7³

or, 7^2x = 7³

Since base is same, power must be equal.

or, 2x = 3

∴ \quad x=\frac{3}{2}=1 \frac{1}{2}

2^x+2 + 2^x−1 = 9

or, 2^x \cdot 2^2+2^x \cdot 2^{-1} = 9

or, 2^x \cdot\left(4+\frac{1}{2}\right) = 9

or, 2^x \cdot 4+\frac{2^x}{2} = 9

or, 2^x \cdot \frac{9}{2} = 9

or, 2^x=2¹

Since base is same, power must be equal.

∴ x = 1

9 × 81^x = 27 ^{2 – x}

or, 3² × 3^4x = 3 ^{3(2 – x)}

or, 3 ^{2 + 4x} = 3 ^{6 – 3x}

Since base is same, power must be equal.

2 + 4x = 6 – 3x

or, 7x = 4

∴ x = 4\over 7

Given: \ 2^{4 x} \cdot 4^{3 x-1}=\frac{4^{2 x}}{2^{3x}}

or, \ 2^{4 x} \cdot 2^{2(3 x-1)}=\frac{2^{2.2 x}}{2^{3 x}}

or, \ 2^{4 x} \cdot 2^{6 x-2}=\frac{2^{4 x}}{2^{3 x}}

or, 2^{4x + 6x – 2} = 2^{4x – 3x}

or, 2^{10x – 2} = 2^x

Since base is same, power must be equal.

10x – 2 = x

or, 9x = 2

∴ x = \frac{2}{9}

Given: 9 × 81^x = 27^{2 – x}

or, 3² × (3⁴)^x = (3³)^{2 – x}

or, 3^{2 + 4x} = 3^{6 – 3x}

Since base is same, power must be equal.

2 + 4x = 6 – 3x

or, 7x = 6 – 2

∴ x = \frac{4}{7}

2 ^{5x + 4} + 2⁹ = 2¹⁰

or, 2 ^{5x + 4} = 2¹⁰ – 2⁹

or, 2 ^{5x + 4} = 2⁹ (2 – 1)

or, 2 ^{5x + 4} = 2⁹

Since base is same, power must be equal.

5x + 4 = 9

or, 5x = 5

∴ x = 1

6^{2x + 4} = 3^3x . 2^{x + 8}

or, (3 × 2)^{2x + 4} = 3^3x . 2^{x + 8}

or, 3^{2x + 4} × 2^{2x + 4} = 3^3x . 2^{x + 8}

Since base is same, power must be equal

2x + 4 = 3x

or, x = 4

(b) 3

Explanation

\ (0.243)^{0.2} \times(10)^{0.6}

= \left(\frac{243}{1000}\right)^{2 \over 10} \times(10)^{6 \over 10}

= \left(\frac{3^{5}}{1000}\right)^{1 / 5} \times 10^{3 \over 5}

= \frac{3^{5 \times {1 \over 5}}}{10^{3 \times {1 \over 5}}} \times 10^{3 \over 5}

= \frac{3^{1}}{10^{3 \over 5}} \times 10^{3 \over 5} = 3

(c) 4

Explanation

2^{1 \over 2} \times 2^{-1\over 2} \times(16)^{1 \over 2} \\

= 2^{{1 \over 2}-{1 \over 2}} \times 2^{4 \times {1 \over 2}}

= 2^{0} \times 2^{2} \\

= 1 × 4 = 4

(b) 9 \over 2

Explanation

4^x = 8³

or,  2^2x = (2³)³

or,  2^2x = 2⁹

Since the base is the same, power must be equal.

or, 2x = 9

or, x = 9 \over 2

20^x = 1\over 7

or, \ \frac{1}{20^{x}}=\frac{1}{7}

or, 20^x = 7

Squaring both sides

or, (20^x)² = 7²

or, (20^x)² = 49

\ 20^{2 x} = (7)^{2} = 49 Ans. (c)

(d) 27

Explanation

4 × 5^x = 500

or, 5^x = 125

or, 5^x = 5³

Since the base is the same, power must be equal.

x = 3

∴ x^x = 3³ = 27

27^x = 81^y

or, 3^3x = 3^4y

Since the base is the same, power must be equal.

3x = 4y

\frac{x}{y} =\frac{4}{3}

∴ x : y = 4 : 3

(ii) \left(5^{5}+0.01\right)^{2}-\left(5^{5}-0.01\right)^{2}=5^x

or, \big\{(5^5+0.01)+(5^5-0.01) \big\} \big\{(5^5+0.01)-(5^5-0.01) \big\} = 5^x

or, (5^5 +0.01)+(5^5-0.01)(5^5 +0.01)-(5^5-0.01)=5^x

or, 2.5^{5} \times 2.01=5^x

or, 2.5^{5} \times 2 \cdot \frac{1}{100}=5^x

or, 5^{5} \times \frac{1}{25}=5^x

or, ^{5} \times \frac{1}{5^{2}}=5^x

or, 5^{5} \times 5^{-2}=5^x

or, 5^{5-2}=5^x

or, 5^{3} = 5^x

∴ x = 3

(iii) 3 \times 27^{x}=9^{x+4},

or, 3 \times 3^{3 x}=3^{2(x+4)}

or, 3^{3 x+1}=3^{2x+8} Since the base is the same, power must be equal.

or, 3x+1=2x+8

or, 3x – 2x = 8-1

or, x = 7

(iv) \sqrt[3]{(1 / 64})^{2/1}

= \left(\frac{1}{64}\right)^{1 / 2^{\times 1 / 3}}

= \left(\frac{1}{64}\right)^{1 / 6}

= \left(\frac{1}{2^{6}}\right)^{1 / 6} \\ =\left(\frac{1}{2}\right)^{6 \times 1 / 6}

= \left(\frac{1}{2}\right)^{1} = 1 / 2 Ans.

(v) 3^{3^{3}},\left(3^{3}\right)^{3}

= 3^{27}

= 3^{3 \times 3}

= 3⁹

∴ 3²⁷ > 3⁹

∴ 3^{3^{3}} is greater