# Chapter – 2 : Laws of Indices | Chapter Solution Class 9

Ganit Prakash 2023 Laws of Indices Solution Class 9. Chapter 2 – Laws of Indices is provided here with simple step-by-step explanations. These solutions for Laws of Indices are extremely popular among class 9 students. The Laws of Indices solution is handy for quickly completing your homework and preparing for exams.

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Laws of Indices (2nd Chapter)

## Let us work out – 2

Question 1

Let us find out the values:

1. (\sqrt[5]{8} )^{5/2}\times (16)^{3/2}
2. \big\{(125)^{-2} \times (16)^ \frac{-3}{2}\big\}^{ \frac{-1}{6}}
3. 4^{1/3}\times [2^{1/3} \times 3^{1/2}] \div 9^{1/4}

Solution:

(i) (\sqrt[5]{8} )^{5/2}×163/2

= (81/5)5/2 × (24)-3/2

= (8)1/5 × 5/2 × (2)4×-3/2

= (23)1/2 × (2)-6

= 23 × 1/2 × 2-6

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

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

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

(ii) \big\{(125)^{-2} \times (16)^ \frac{-3}{2}\big\}^{ \frac{-1}{6}}

= \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 Ans.

(iii) 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

Question 2

Let us Simplify:

(i) (8a^3 \div 27x^{-3})^\frac{2}{3} \times (64a^3 \div 27x^{-3})^\frac{2}{3}

(ii) \big\{(x^{-5})^{2/3}\big\}^{-3/10}

(iii) [ \big\{(2^{-1})^{-1}\big\} ^{-1}]^{-1}

(iv) \sqrt[3]{a^{-2}} . b \times \sqrt[3]{b^{-2}} .c \times \sqrt[3]{c^{-2}} . a

(v) \frac{4^{m+\frac{1}{4}} \times \sqrt{2.2^m}^{\frac{1}{m}}}{2. \sqrt{2^{-m}} }

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

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

Solution:

(i) (8a^3 \div 27x^{-3})^\frac{2}{3} \times (64a^3 \div 27x^{-3})^\frac{2}{3}

= (\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} Ans.

(ii) \big\{(x^{-5})^{2/3}\big\}^{-3/10}

= \big\{x^{-5 \times 2/3}\big\}^{-3/10}

= \big\{x^{-10/3}\big\}^{-3/10}

= x^{-10/3x-3/10}

= x^1

= x

(iii) [ \big\{(2^{-1})^{-1}\big\} ^{-1}]^{-1}

= [ \big\{(2^{-1})^{-1}\big\} ^{-1}]^{-1}

= [2^{1 \times -1}]^{-1}

= 2^{-1 \times -1}

= [\big\{2^1\big\}^{-1}]^{-1}

= [2^{-1 \times -1}]

= [2^-1]

= 2 Ans.

(iv) \sqrt[3]{a^{-2}} . b \times \sqrt[3]{b^{-2}} .c \times \sqrt[3]{c^{-2}} . a

= a^{-2/3} . b \times b^{-2/3} .c \times c^{-2/3} . a

= a^{-2/3} . a \times b^{-2/3} .b \times c^{-2/3} . c

= a^{-2/3+1} . a \times b^{-2/3+1} .b \times c^{-2/3+1} . c

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

(v) \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} }

= 2^3

= 8 Ans.

(vi) 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.

(vii) (\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 b+b^2} \times\left(X^{b-c}\right)^{b^2+b c+c^2} \times\left(X^{c-a}\right)^{c^2+c a+a^2} \\

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

Question 3

Let us Arrange in ascending order:

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

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

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

Solution:

(i) 5^{1 / 2}, 10^{1 / 4}, 6^{1 / 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

\left(5^{1 / 2}\right)^{12 / 12}=\left(5^{1 / 2^{\times 12}}\right)^{1 / 12}=\left(5^6\right)^{1 / 12}=(15,625)^{1 / 12} \\ \left(10^{1 / 4}\right)^{12 / 12}=\left(10^{1 / 4} \times 12\right)^{1 / 12}=\left(10^3\right)^{1 / 12}=(1000)^{1 / 12} \\ \left(6^{1 / 3}\right)^{12 / 12}=\left(6^{1 / 3^* \times 12}\right)^{1 / 12}=\left(6^4\right)^{1 / 12}=(1296)^{1 / 12}

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

(ii) 3^{1 / 3}, 2^{1 / 2}, 8^{1 / 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

\left(3^{1 / 3}\right)^{12 / 12}=\left(3^{1 / 3} \times 12\right)^{1 / 12}=\left(3^4\right)^{1 / 12}=(81)^{1 / 12} \\

\left(2^{1 / 2}\right)^{12 / 12}=\left(2^{1 / 2^{\times 12}}\right)^{1 / 12}=\left(2^6\right)^{1 / 12}=(64)^{1 / 12} \\

\left(8^{1 / 4}\right)^{12 / 12}=\left(8^{1 / 4^{\times 12}}\right)^{1 / 12}=\left(8^3\right)^{1 / 12}=(512)^{1 / 12} \\

∴ In ascending order, 2^{1 / 2}, 3^{1 / 3}, 8^{1 / 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} Ans.

Question 4

Let us Prove:

{ (i) }\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=1

{ (ii) }\left(\frac{\mathrm{X}^{\mathrm{m}}}{\mathrm{X}^{\mathrm{n}}}\right)^{\mathrm{m}-\mathrm{n}} \times\left(\frac{\mathrm{X}^{\mathrm{n}}}{\mathrm{X}^{\prime}}\right)^{\mathrm{n}+1} \times\left(\frac{\mathrm{X}^l}{\mathrm{X}^{\mathrm{m}}}\right)^{l+\mathrm{m}}=1

{ (iii) }\left(\frac{X^{\mathrm{m}}}{\mathrm{X}^{\mathrm{n}}}\right)^{\mathrm{m}+\mathrm{n}+l}\left(\frac{\mathrm{X}^{\mathrm{n}}}{\mathrm{X}^l}\right)^{\mathrm{n}+l+\mathrm{m}} \times\left(\frac{X^l}{\mathrm{X}^{\mathrm{m}}}\right)^{l+\mathrm{m}+\mathrm{m}}=1

{ (iv) }\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

Solution:

(i) 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 \quad \therefore \text { L.H.S = R.H.S. }

(ii) \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^{P^2-m^2} \\

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

=X^0=1 \therefore \text { L.H.S = R.H.S. (proved) }

(iii) \left(\frac{X^m}{X^n}\right)^{m+n+l} \times\left(\frac{X^n}{X^I}\right)^{n+I+m}\left(\frac{X^{\prime}}{X^m}\right)^{I+m+n} \\ =\left(X^{\mathrm{m}-\mathrm{n}}\right)^{\mathrm{m}+\mathrm{n}+1} \times\left(\mathrm{X}^{\mathrm{n}-l}\right)^{\mathrm{n}+1+\mathrm{m}} \times\left(\mathrm{X}^{1-\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. \\ \left(X^{\left.m^2+m p+l m-m n-n^2-l n\right)} \times X^{\left.n^2+l p^{\prime}+m n-l p^2-l^2-l m\right)}\right.\times \mathrm{X}^{\left.l^2+l m+l n-l m-m^2-m n\right)}\\ =\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} \\ X^0=1 \therefore \text { L.H.S = R.H.S. (proved) } \\

(iv) \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)}} \\

=a^0=1 \therefore \text { L.H.S = R.H.S. (proved) } \\

Question 5

If x + z = 2y and b2 = ac, then let us show that a^{y-z}b^{z-x}c^{x-y} = 1

Solution:

We have,

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

xy = yz…………….(i)

L.H.S.=a^{y-z} \cdot b^{z-x} \cdot c^{y-z} \\

=a^{y-z} \cdot b^{z-x} \cdot c^{y-z}[b y(i)] \\

=(ac)^{y-z} \cdot b^{z-x} \\

=\left(b^2\right)^{y-z} \cdot b^{z-x} \quad\left[\because b^2=a c\right] \\

=b^{2 y-2 z} \cdot b^{z-x} \\

= b^{2 y-2 z+z-x}=b^{2 y-x-z}=b^{x+z-x-z}[\because x+z=2 y] \\

= b^0 = 1 (Ans)

Question 6

If a = xyp-1, b = xyq-1 and c = xyr-1 then let us show that a^{q-r}b^{r-p}c^{p-q} = 1

Solution:

{ L.H.S. } \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 \therefore \text { L.H.S = R.H.S. (proved) }

Question 7

If x^{\frac{1}{a}} = y^{\frac{1}{b}} = z^{\frac{1}{b}} and xyz = 1, then let us show that a + b + c= 0

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

∴ x = ka, y = kb. z = kc

and xyz = 1

or, ka .kb .kc = 1

or, ka+b+c = K0

Since base is same, power must be equal.

a + b + c = 0 (Proved)

Question 8

If ax=by=cz and abc = 1, then let us show that xy + yz + zx = 0

Solution:

Let ax= by= c= k(Say)

a=k^{1 / x}, b=k^{1 / y}, c=k^{1 / z}\\

\therefore \quad a b c=1 \\

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

\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k^0\left[\because k^0=1\right]

Since base is same, power must be equal.

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

\therefore \quad x y+y z+z x=0 \text { (Proved) }

Question 9

Let us solve:

(i) 49^x=7^3

(ii) 2^{\mathrm{x}+2}+2^{\mathrm{x}-1}=9

(iii) 2^{\mathrm{x}+1}+2^{\mathrm{x}+2}=48

(iv) 2^{4 x} \cdot 4^{3 x-1}=\frac{4^{2 x}}{2^{3 x}}

(v) 9 \times 81^x=27^{2-x}

(vi) 2^{5 x+4}+2^9=2^{10}

(vii) 6^{2 x+4}=3^{3 x} \cdot 2^{x+8}

Solution:

(i) 49^x=7^3

or, 7^{2 x}=7^3

Since base is same, power must be equal.

{ or, } 2 x=3 \\

\therefore \quad x=\frac{3}{2}=1 \frac{1}{2} \text { (Ans.) }

(ii) 2^{\mathrm{x}+2}+2^{\mathrm{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=2^1

Since base is same, power must be equal.

∴ x = 1 (Ans.)

(iii) 2^{\mathrm{x}+1}+2^{\mathrm{x}+2}=48 \

or, \ 2^{x} \cdot 2+2^{x} \cdot 2^{2}=48

0r, \ 2^{x} \cdot\left(2+2^{2}\right)=48 \

or, \ 2^{x} \cdot 6=48 \

or, \ 2^{x}=\frac{48}{6} \

or, \ 2^{x} .=8 \

or, \ 2^{x} .=2^{3}\

Since base is same, power must be equal.

(iv) \ 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^{4 x+6 x-2}=2^{4 x-3 x}

or, \ 2^{10 x-2}=2^{\mathrm{x}} Since base is same, power must be equal.

\\ 10 x-2=x \\

\text or, 10 x-x=2

(v) \ 9 \times 81^{x}=27^{2-x}

or, \ 3^{2}\times\left(3^{4}\right)^{x}=\left(3^{3}\right)^{2-x}

or, ( 3^{2} \times\left(3^{4 x}\right)=3^{3(2-x)}

or, \ 3^{2+4 x}=(3)^{6-3 x}

\ \therefore \quad 2+4 x=6-3 x Since base is same, power must be equal.

or, 4x + 3x = 6-2

or, 7x = 4

\ \therefore x = \frac{4}{7}

(iv) \ 2^{5 x+4}+2^{9}=2^{10}

or, \ 2^{5 x+4}=2^{10}-2^{9}

or, \ 2^{5 x+4}=2^{9}(2-1)

or, \ 2^{5 x+4}=2^{9}

Since base is same, power must be equal.

5x + 4 = 9

or, \ 5 x=9-4

or, \ 5 x=5

or, \ x=\frac{5}{5}

(vii) \ 6^{2 x+4}=3^{3 x} \cdot 2^{x+8}

or, \ (3.2)^{2 x+4}=3^{3 x} \cdot 2^{x+8}

or, \ 3^{2 x+4} \cdot 2^{2 x+4}=3^{3 x} \cdot 2^{x+8}

or, \ \frac{2^{2 x+4}}{2^{x+8}}=\frac{3^{3 x}}{3^{2 x+4}}

or, \ 2^{2 x+4-(x+8)}=3^{3 x-(2 x+4)}

or, \ 2^{2 x+4-x-8}=3^{3 x-2 x-4}

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

or, \ \frac{2^{x-4}}{3^{x-4}}=1

or, \ \left(\frac{2}{3}\right)^{x-4}=\left(\frac{2}{3}\right)^{0}

Since base is same, power must be equal.

or, x - 4 = 0 or, x = 4 (Ans)

Question 10

Multiple Choices Question (M.C.Q):

(i) The value of \ (0.243)^{0.2} \times(10)^{0.6} is

(a) 0.3

(b) 3

(c) 0.9

(d) 9

Solution:

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

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

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

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

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

= 3 Ans. (b)

(ii)  The value of \ 2^{1 / 2} \times 2^{-1 / 2} \times(16)^{1 / 2} is

(a) 1

(b) 2

(c) 4

(d) \frac{1}{2}

Solution:

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

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

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

= 1 \times 4=4  Ans.(c)

(iii) If \ 4^{x}=8^{3} then the value of x is

(a) \ 3 / 2

(b) 9 / 2

(c) 3

(d) 9

Solution:

4^{x}=8^{3}

or, \ 2^{2 x}=2^{3.3}

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

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

or, \ 2{x}=9

or, \ x=9 / 2 Ans.(b)

(iv) If  \ 20^{-x}=1 / 7, then the value of \ (20)^{2 x} is

(a) \ 1 / 49

(b) 7

(c) 49

(d) 1

Solution:

20^x = 1 / 7

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

or, \ 20^{x}=7, Squaring both sides.

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

(v) If \times 5^{x} = 500, then the value of x^{x} is

(a) 8

(b) 1

(c) 64

(d) 27

Solution:

4 \times 5^x = 500

or, 5^x =\frac{500}{4}

or, 5^x = 125 or, 5^x = 5^3

x = 3 \quad\therefore x^3 = 3^3 = 27 Ans. (d)

Question 11

(i) If (27)^{x} = (81)^{y}, then let us write x : y

(ii) \left(5^{5}+0.01\right)^{2}-\left(5^{5}-0.01\right)^{2} = 5x, then let us calculate the value of x and write it.

(iii) If 3 \times 27^{x} = 9^{x+4}, then let us calculate the value of x and write it.

(iv) Let is find out the value of \ \sqrt[3]{(1 / 64})^{1/2} and write it.

(v) Let us write explaining the greater value between 3^{3^{3}} and \left(3^{3}\right)^{3} with reason

Solution:

(i) (27)^{x} = (81)^{y}

or, \left(3^{3}\right)^{x}=\left(3^{4}\right)^{y}

3^{3 x}=3^{4 y}

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

3x=4y

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

\therefore \quad x: y =4 : 3 (Ans.)

(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 \therefore 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, 3 x-2 x=8-1

or, x = 7 Ans.

(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}

= 3^{9} \therefore 3^{27}>3^{9} \therefore 3^{3^{3}} is greater

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