Ganit Prakash 2024 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:
(i) (\sqrt[5]{8} )^{5\over2}\times (16)^{3\over2}
Solution:
(\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}}
Solution:
= \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
(iii) 4^{1\over3}\times [2^{1\over3} \times 3^{1\over2}] \div 9^{1\over4}
Solution:
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}
Solution:
= (\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}
(ii) \big\{(x^{-5})^{2/3}\big\}^{-3/10}
Solution:
= \big\{x^{-5 \times {2\over3}}\big\}^{-3\over10}
= \big\{x^{-10\over3}\big\}^{-3\over10}
= x^{-{10\over3x}×{-3\over10}} = x
(iii) [{(2-1)-1}-1]-1
Solution:
[{(2-1)-1}-1]-1
= {(2)-1}-1
= 2
(iv) \sqrt[3]{a^{-2}} . b \times \sqrt[3]{b^{-2}} .c \times \sqrt[3]{c^{-2}} . a
Solution:
= 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}}
(v) \frac{4^{m+\frac{1}{4}} \times \sqrt{2.2^m}^{\frac{1}{m}}}{2. \sqrt{2^{-m}} }
Solution:
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
(vi) 9^{-3} \times \frac{16^{\frac{1}{4}}}{6^{-2}} \times ( \frac{1}{27})^{- \frac{4}{3}}
Solution:
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.
(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:
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 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 \over 2}, 10^{1 \over 4}, 6^{1 \over 3}
(ii) 3^{1 \over 3}, 2^{1 \over 2}, 8^{1 \over 4}
(iii) 2^{60}, 3^{48}, 4^{36}, 5^{24}
Solution:
(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
\left(5^{1 \over 2}\right)^{12 \over 12}=\left(5^{1 / 2^{\times 12}}\right)^{1 \over 12}=\left(5^6\right)^{1 \over 12}=(15,625)^{1 \over 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 \over 3}\right)^{12 \over 12}=\left(6^{1 \over 3^* \times 12}\right)^{1 \over 12}=\left(6^4\right)^{1 \over 12}=(1296)^{1 \over 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
\left(3^{1 \over 3}\right)^{12 \over 12}=\left(3^{1 \over 3} \times 12\right)^{1 \over 12}=\left(3^4\right)^{1 \over 12}=(81)^{1 \over 12} \\ \left(2^{1 \over 2}\right)^{12 \over 12}=\left(2^{1 \over 2^{\times 12}}\right)^{1\over12}=\left(2^6\right)^{1 / 12}=(64)^{1 \over 12} \\ \left(8^{1 \over 4}\right)^{12 \over 12}=\left(8^{1 / 4^{\times 12}}\right)^{1 \over 12}=\left(8^3\right)^{1 / 12}=(512)^{1 \over 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}
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
Solution:
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
(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
Solution:
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)
(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
Solution:
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)
(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:
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)
Question 5
If x + z = 2y and b2 = ac, then let us show that a y – z bz – x cx – y = 1
a^{y-z}b^{z-x}c^{x-y} = 1
Solution:
We have,
x + z = 2y or x + z = y + y
x – y = y – z …… (i)
L.H.S: a y – z bz – x cx – y
= a y – z bz – x cy – z
= (ac)y – z bz – x
= (b2)y – z bz – x (since b2 = ac)
= b2y – 2z bz – x
= b 2y – 2z + z – x
= b 2y – (x + z)
= b 2y – 2y = b0 = 1 : RHS
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:
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)
Question 7
If x^{\frac{1}{a}} = y^{\frac{1}{b}} = z^{\frac{1}{c}} 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}{c}} = 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= cz = k (Say)
a = k1/x ; b = k1/y ; c = k1/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)
Question 9
Let us solve:
(i) 49^x=7^3
Solution:
Given: 49x = 73
or, 72x = 73
Since base is same, power must be equal.
or, 2x = 3
∴ \quad x=\frac{3}{2}=1 \frac{1}{2}
(ii) 2x+2 + 2x−1 = 9
Solution:
2x+2 + 2x−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, 2x=21
Since base is same, power must be equal.
∴ x = 1
(iii) 9 × 81x = 27 2 – x
Solution:
9 × 81x = 27 2 – x
or, 32 × 34x = 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
(iv) 2^{4 x} \cdot 4^{3 x-1}=\frac{4^{2 x}}{2^{3 x}}
Solution:
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, 24x + 6x – 2 = 24x – 3x
or, 210x – 2 = 2x
Since base is same, power must be equal.
10x – 2 = x
or, 9x = 2
∴ x = \frac{2}{9}
(v) 9 × 81x = 272 – x
Solution:
Given: 9 × 81x = 272 – x
or, 32 × (34)x = (33)2 – x
or, 32 + 4x = 36 – 3x
Since base is same, power must be equal.
2 + 4x = 6 – 3x
or, 7x = 6 – 2
∴ x = \frac{4}{7}
(vi) 2 5x + 4 + 29 = 210
Solution:
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
(vii) 62x + 4 = 33x . 2x + 8
Solution:
62x + 4 = 33x . 2x + 8
or, (3 × 2)2x + 4 = 33x . 2x + 8
or, 32x + 4 × 22x + 4 = 33x . 2x + 8
Since base is same, power must be equal
2x + 4 = 3x
or, x = 4
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:
(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
(ii) The value of \ 2^{1 \over 2} \times 2^{-1 \over 2} \times(16)^{1 \over 2} is
(a) 1
(b) 2
(c) 4
(d) \frac{1}{2}
Solution:
(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
(iii) If \ 4^{x}=8^{3} then the value of x is
(a) 3 \over 2
(b) 9 \over 2
(c) 3
(d) 9
Solution:
(b) 9 \over 2
Explanation
4x = 83
or, 22x = (23)3
or, 22x = 29
Since the base is the same, power must be equal.
or, 2x = 9
or, x = 9 \over 2
(iv) If \ 20^{-x}={1 \over 7}, then the value of \ (20)^{2 x} is
(a) 1 \over 49
(b) 7
(c) 49
(d) 1
Solution:
20x = 1\over 7
or, \ \frac{1}{20^{x}}=\frac{1}{7}
or, 20x = 7
Squaring both sides
or, (20x)2 = 72
or, (20x)2 = 49
\ 20^{2 x} = (7)^{2} = 49 Ans. (c)
(v) If 4 \times 5^{x} = 500, then the value of xx is
(a) 8
(b) 1
(c) 64
(d) 27
Solution:
(d) 27
Explanation
4 × 5x = 500
or, 5x = 125
or, 5x = 53
Since the base is the same, power must be equal.
x = 3
∴ xx = 33 = 27
Question 11
Short answer type question:
(i) If 27x = 81y, then let us write x : y
Solution:
27x = 81y
or, 33x = 34y
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} = 5x, then let us calculate the value of x and write it.
Solution:
(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) If 3 \times 27^{x} = 9^{x+4}, then let us calculate the value of x and write it.
Solution:
(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) Let is find out the value of \ \sqrt[3]{(1 / 64})^{1/2} and write it.
Solution:
(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) Let us write explaining the greater value between 3^{3^{3}} and \left(3^{3}\right)^{3} with reason
Solution:
(v) 3^{3^{3}},\left(3^{3}\right)^{3}
= 3^{27}
= 3^{3 \times 3}
= 39
∴ 327 > 39
∴ 3^{3^{3}} is greater