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(a) 103

Given exp. = \(\frac{1}{{(6^3)}^{-\frac{2}{3}}}+\frac{1}{(4^4)^{-\frac{3}{4}}}+\frac{1}{(3^5)^{-\frac{1}{5}}}\)

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

= 6² + 4³ + 3 = 36 + 64 + 3 = 103

Correct option is A) 1

(-1)¹⁰⁶ = 1

(\(\because\) (-1)^even = 1 & 106 is an even number)

By the law of exponents \((\frac{a}{b})^0=1\)

We will get,

\((\frac{3}{5})^0=1\)

Correct option is C) 0

(4) (4° - 3°) x 2° = (1 - 1) x 1 = 0 x 1 = 0

Correct option is D) y³ × y⁵ = y⁸

y³ x y⁵ = y^{3 + 5} = y⁸

\((\frac{5}{12})^{-4}\times(\frac{5}{12})^{3x}=(\frac{5}{12})^5\)

\(\Rightarrow (\frac{5}{12})^{-4+3x}=(\frac{5}{12})^5\)

\(\Rightarrow\) 3x = 5 + 4 = 9

\(\Rightarrow x=\frac{9}{3}=3\)

Correct option is C) -5

7^x+5 = 1 = 7⁰

⇒ x + 5 = 0

⇒ x = -5

(-27/64)

We have,

-27 = -3 × -3 × -3 = (-3)³

64 = 4 × 4 × 4 = (4)³

Then,

= (-3³/4³)

∴ (-3/4)³

\(3^{-6}\div3^4=(\frac{1}{3^6}\div3^4)\)

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

\(=\frac{1}{3^{10}}=3^{-10}\)

\((\frac{3}{4})^{-3}=(\frac{4}{33})^3=\frac{4^3}{3^3}=\frac{64}{27}\)

(i) Given 4.83 × 10⁷ 

4.83 × 10⁷ = 483 × 10^7-2 [since the decimal point is moved two places to the right]

= 483 × 10⁵ 

= 4, 83, 00,000

(ii) Given 3.21 × 10⁵ 

3.21 × 10⁵ = 321 x 10^5-2 [since the decimal point is moved two places to the right]
= 321 x 10³ 

= 3, 21,000

(iii) Given 3.5 × 10³ 

3.5 × 10³ = 35 x 10^3-1 [since the decimal point is moved one place to the right]

= 35 x 10² 

= 3,500

(i) 3⁷ × 3³ = 3⁷ + 3³ = 3¹⁰       [∵ a^m × aⁿ = a^m+n]

(ii) 4 × 4 × 4 × 4 × 4 = 4⁵      [∵ a × a × a × ……. m times = a^m]

(iii) 3⁴ × 4³ = 3⁴⁺³ = 3⁷      [∵ a^m × aⁿ = a^m+n]

\((\frac{-1}{3})^3=\frac{-1^3}{3^3}=\frac{-1}{27}\)

[{7⁴} ^-1/3]^1/4

= (1/7⁴)^{1/3 × 1/4}

= (1/7)^1/3 = 7^m

= 7^-1/3 = 7^m

= m = -1/3

\(\{m\sqrt{n\sqrt{a}}\}^{mn}\)

We know for any non-zero number a,

a^m × aⁿ = a^m+n

= ^{\(\{(a^{\frac{1}{n}})^{\frac{1}{m}}\}^{mn}\)}

Again using (a^m)ⁿ = a^mn we get, = \(\{a^{\frac{1}{mn}}\}^{mn}\) = a

(2/4)^{2 – 4} + (4 / 2)^{4 – 2}

= (1/2)^-2 + 2²

= 2² + 2²

= 8

(i) \((\frac{\sqrt1}{\sqrt3})^5\) = (1 / x^3/2) 5

= (1 / x ^{3/2 × 5}) = (1 / x ^15/2)

(ii) \((\sqrt3^3/\sqrt{y}^2)\) = (x³ / y²) ^1/2

= x^{3 × 1/2} / y^{2 × 1/2} 

= x^3/2 / y

(iii) 1 / (x^2/3 y^{1/2₎ 2} 

= 1 / (x^{2/3 × 2} y^{1/2 × 2}) 

= 1 / x^4/3 y

(iv) (x^1/2) ^-2/3 (y)² / (xy^-1/2) ^1/2

= x^-1/3y ² / (x^1/2y ^-1/2×1/2)

= (x^-5/6) (y^9/4) 

= (y^9/4) / (x^5/6)

(v) (243x¹⁰ y⁵ z¹⁰) ^1/5 

= (3⁵) ^1/5 x²yz² 

= 3x²yz² 

(vi) (y¹⁰ / x⁴) ^5/4 

= y ^{10 × 5/4} / x^{4 × 5/4} 

= y^25/2 / x⁵

We have, 

(27)^x = 9 / 3^x 

(3³) ^x = 3² / 3^x 

3 ^3x = 3 ^{2 – x} 

3x = 2 – x {On equating exponents} 

3x + x = 2 

4x = 2 

x = \(\frac{2}{4}\) \(=\frac{1}{2}\)

Hence, the value of x is \(\frac{1}{2}\)

(i) 1/2 + 1/(0.01) ^1/2 -3² 

=1/2 + 10 – 9 

=1/2 + 1 

=3/2 

(ii) (2 ⁿ + 2^n-1)/ ) (2ⁿ⁺¹ - 2ⁿ) 

=2 ⁿ(1 + 2^-1 ) / 2ⁿ (2-1) 

= [1 + (1/2)]/1 

=1 + 1/2 

=3/2 

(iii) (125/64) ^2/3 + (625/256)^1/4 + ( 5/4) 

=(5/4) ² + 5/4 + 5/4 

=25/16 + 5/4 + 5/4 

=65/16 

(iv) (3 ^-3.6².7(2)^1/2)/ (5^4/3.(15)^-4/3‑.3^1/3) 

=28(2)^1/2 (3 ^-3.36.7(2)^1/2)/ (5^4/3-4/3.(3)^-1) 

(3 ^-2.36.7(2)^1/2)/ (5⁰) 

1/9.36.7(2) ^1/2 

28\(\sqrt2\)

(v) {1- 1/0.1}/ { (3/8) ^-1(3/2)³ + (-1/3)^-1 

=1-10/{ (8/3)(3/2) ³ + (-3) 

=-9/(3 ²-3) 

= -3/2

(3125a¹⁰b⁵c¹⁰) ^1/5

= 5a²bc²

x ² = 2

x = \(\sqrt2\)

x ³ = (2)^{1/2 × 3}

= 2\(\sqrt2\)

(x)^ab × 1/^ab . (x)^bc . 1/bc . x^ca . 1/ca 

= x . x . x 

= x³

5ⁿ (25 – 30) / 5ⁿ (13 – 10)

= -5 / 3

\(\frac{{3}^{2x-8}}{225}\)= \(\frac{5^3}{5^x}\)

= 5^x × 3^2x – 8 = 5⁵ × 3³ Comparing the coefficient of x we get,

= x = 5

(i) (3^1/2+1/6.5^{-3/2 +1}) / (3^-1/3.5^1/2) 

=(3 ^2/3.5^-1/2) / (3^-1/3.5^1/2) 

=(3 ^2/3 + ^1/3) / (5^{1/2 +1/2}) 

=3/5 

(ii) (3 ² )^3/2 -3.1 – (1/9²) ^-1/2 

= 3 ³ -3 -9 

=27 -3 -9 

=27-12 

=15 

(iii) 2 ^(-2)(-2) -3.8^2/3 +(3/4)^-1 

=2 ⁴ -3.2² + 4/3 =16 -12 + 4/3 

=16/3 

(iv) [(2.3 ^1/3)/(2^-1/5 5^2/5)] × (2^-1/5.3)/ (3^4/3.5^7/5) 

= 2.3 ^{1/3 +1 -4/3} / 5^2/5-7/5 

= 2.5 

=10

(b/a)^1/2 × (c/b)^1/2 × (a/c)^1/2

= (b/a × c/b × a/c)^1/2

= 1

Correct option is B) 2⁴ × 3⁴

(i) Given (25)³ ÷ 5³

= (5²)³ ÷ 5³[According to the law of exponents we have (a^m)ⁿ = a^mn]

= 5⁶ ÷ 5³ [According to the law of exponents we have a^m ÷ aⁿ = a^m-n]         

= 5^{6 – 3}

= 5³

(ii) Given (81)⁵ ÷ (3²)⁵[According to the law of exponents we have (a^m)ⁿ = a^mn]

= (81)⁵ ÷ 3¹⁰[81 = 3⁴]

= (3⁴)⁵ ÷ 3¹⁰ [According to the law of exponents we have (a^m)ⁿ = a^mn]

= 3²⁰ ÷ 3¹⁰

= 3^20-10 [According to the law of exponents we have a^m ÷ aⁿ = a^m-n]         

= 3¹⁰

(iii) Given 9⁸ × (x²)⁵/ (27)⁴ × (x³)²

= (3²)⁸ × (x²)⁵/ (3³)⁴× (x³)²[According to the law of exponents we have (a^m)ⁿ = a^mn]

= 3¹⁶ × x¹⁰/3¹² × x⁶

= 3^16-12 × x^10-6[According to the law of exponents we have a^m ÷ aⁿ = a^m-n]         

= 3⁴ × x⁴

= (3x)⁴

(iv) Given (3² × 7⁸ × 13⁶)/ (21² × 91³)

= (3² × 7²7⁸ × 13⁶)/(21²× 13³ × 7³)[According to the law of exponents we have (a^m)ⁿ = a^mn]

= (21² × 7² × 13⁶)/(21²× 13³ × 7³)

= (7⁶ × 13⁶)/(13³ × 7³)

= 91⁶/91³[According to the law of exponents we have a^m ÷ aⁿ = a^m-n] 

= 91^6-3

= 91³

(i) 4³ or 3⁴

4³ = 4 × 4 × 4 = 64

3⁴ = 3 × 3 × 3 × 3 = 81

∴ 3⁴ > 4³

(ii) 5³ or 3⁵

5³ = 5 × 5 × 5 = 125

3⁵ = 3 × 3 × 3 × 3 × 3 = 243

∴ 3⁵ >5³

(iii) 2⁸ or 8²

2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8² = 8 × 8 = 64

∴ 2⁸ > 8²

(iv) 100² or 2¹⁰⁰

100² = 100 × 100= 10,000

2⁸ = (2¹⁰)¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
× 2 × 2

= (1024)¹⁰ = [(1024)²]⁵ = [10,48,576]

= (10,48,476)⁵ > 10,000

= 2¹⁰⁰ > 100²

(v) 2¹⁰ or 10²

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

10² = 10 × 10 = 100

∴ 2¹⁰ > 10²

(4/25)^m = (2/5)^{m^2}

(2²/5²)^m = (2/5)^m^²

(2/5)^2m = (2/5)^{m^2}

∴ m² = 2m 

m.m = 2m 

m = 2 

It can happen if m = 2

\(\frac{3^m \times2^n}{2^m\times3^n}\) = \(\frac{3^{m - n}}{2^{m - n}}\)

If ‘m’ and ‘n’ are two distinct integers. 

Then 3^(m-n) is always on odd number and 2^(m-n) is always an even number. If odd number is divided, by an even number the quotient is not an integer. 

Therefore it is’not an integer.

2⁴ = 2 × 2 × 2 ×2 = 16;

2³ = 2 × 2 x 2 = 8

2⁷ = 2 × 2 × 2 × 2 × 2 × 2  × 2 = 128

2⁴ × 2³ = 16 × 8 = 128 = 2⁷

2⁴ × 2³ = 2⁷

(i) 3² × 3⁴ × 3⁸
= 3^{2 + 4 + 8} = 3¹⁴ [∵ a^m × aⁿ = a^{m + n}]

(ii) 6¹⁵ ÷ 6¹⁰

= 6^{15 – 10} = 6⁵ [∵ a^m ÷ aⁿ = a^{m – n}]

(iii) a³ × a²

= a^{3 + 2} = a⁵ [∵ a^m × aⁿ = a^{m + n}]

(iv) 7^x × 7²

= 7^{x + 2} [∵ a^m × aⁿ = a^{m + n}]

(v) (5²)³ ÷ 5³

= 5^{2 × 3} ÷ 5³ = 5^{6 – 3} [(a^m)ⁿ = a^mn

∵ a^m ÷ aⁿ = a^{m – n}]

(vi) 2⁵ × 5⁵

= (2 × 5)⁵ = 10⁵ [∵ a^m × b^m = (ab)^m]

(vii) a⁴ × b⁴

= a⁴ × b⁴ = (ab)⁴ [∵ a^m × b^m = (ab)^m]

(viii) (3⁴)³

= 3^{4 × 3} = 3¹² [(a^m)ⁿ = a^mn]

(ix) (2²⁰ ÷ 2¹⁵)

(2²⁰ ÷ 2¹⁵) × 2³ = 2^{20 – 15} × 2³ [a^m ÷ aⁿ = a^{m – n}]

= 2⁵ × 2³

= 2^{5 + 3} = 2⁸ [a^m × aⁿ = a^{m + n}]

(x) 8^t ÷ 8²

= 8^{t – 2} [∵ a^m ÷ aⁿ = a^{m – n}]

(i) 10 × 10¹¹ = 100¹¹

L.H.S = 10 × 10¹¹ = 10^{1 + 11} = 10¹² RHS

100¹¹ = (10 × 10)¹¹

= (10²)¹¹

= 10² × 11 = 10²²

∴ 10¹² ≠ 10²²

So 10 × 10¹¹ ≠ 100¹¹ 

∴ It is false.

(ii) 2³ > 5²

= 2 × 2 × 2 > 5 × 5

= 8 > 25

∴ 2³ > 5² 

∴ It is false.

(iii) 2³ × 3² = 6⁵

= 2 × 2 × 2 × 3 × 3

= 6 × 6 × 6 × 6 × 6

= 72 = 7776

∴ 2³ × 3² ≠ 6⁵ 

∴ It is false.

(iv) 3⁰ = (1000)⁰

1 = 1

∴ 3⁰ = (1000)⁰

 ∴ It is true

The quotient law of exponent states that to divide two exponents with the same base, you keep the base and subtract the powers.

3² ≠ 2³

Since 3² = 3 × 3 = 9 and 2³ = 8

∴ 3² ≠ 2³

i) 51 × 10⁷ = 3× 17 × 10⁷

ii) 7 × 10⁷

iii) 4550 millions = 4550 × 10,00,000 (v 1 million =10 lakhs)

= 455 × 10⁷ = 91 × 5 × 10⁷ = 5 × 7 × 13 × 10⁷

iv) 1 km = 1000 m

∴ 1000 km = 1000 × 1000 m = 10⁶

i) 48951 = (4 × 10000) + (8 × 1000) + (9 × 100) + (5 × 10) + (1 × 1)

= (4 × 104) + (8× 103) + (9 × 102) + (5 × 1.0) + (1 × 1)

ii) 89325 = (8 × 10000) + (9 × 1000) + (3 × 100) + (2 × 10) + (5 × 1)

= (8 × 10⁴) + (9 × 10³) + (3 × 10²) + (2 × 10) + (5 × 1)

(i) 345 = 3.45 × 100 = 3.45 × 10²

(ii) 180000 = 18 × 1000 = 18 × 10⁴ = 1.8 × 10 × 10⁴ =1.8 × 10⁽¹⁺⁴⁾ =1.8 × 10⁵

(iii) 0.000003 = \(\frac{3}{1000000}\) = 3 × 10^-6

(iv) 0.000027 = \(\frac{27}{1000000}=\frac{27}{10^6}=\frac{2.7\times10}{10^6}\) = 2.7 × 10^(1-6) = 2.7 × 10^-5

2⁰ + 3⁰ = 1 + 1 = 2 

[∵ a⁰ = 1]

Population of India in March 2001= 1027000000

A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.

Then,

Population of India in March 2001= 1027000000

= (1.027 × 10⁹) (in standard form)

Distance between Earth and Moon = 384000000 m

A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.

Then,

Distance between Earth and Moon = 384000000 m

= (3.84 × 10⁸) m (in standard form)

g = t ^2/3 + 4t ^-1/2 

= (64)^2/3 + 4 (64) ^-1/2

= [(64)^1/3] ³ + 4 \((\cfrac{1}{64})^\cfrac{1}{2}\)

= 4² + 4 (\(\cfrac{1}8\) )

= 16 + \(\cfrac{1}2\) = \(\cfrac{33}2\)

\((\cfrac{1}{8})^2\) = (8)²

\(\cfrac{1}{x}\) = 8

x = \(\cfrac{1}{8}\)

x^1/3 + x^o

= \((\cfrac{1}{8})^{\cfrac{1}{3}}\) + \((\cfrac{1}{8})^0\)

= \(\cfrac{1}{2} + 1\) = \(\cfrac{3}{2}\)

(d) \(\frac{11}{96}\)

2^x = (2²)^y = (2³)^z ⇒ x = 2^y = 3^z

Given xyz = 288 ⇒ \(x\times\frac{x}{2}\times\frac{x}{3} = 288\)

⇒ x³ = 6 × 288 ⇒ x³ = 1728

⇒ x = \(\sqrt[3]{1728}=12\)

∴ y = \(\frac{12}{2}\) = 6 and  = \(\frac{12}{3}\) = 4

∴ \(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}\) = \(\frac{1}{24}+\frac{1}{24}+\frac{1}{32}\)

= \(\frac{4+4+3}{96} = \frac{11}{96}.\)

(2.4)^x = 10z ⇒ 2.4 = \(10^{\frac{z}{x}}\) 

and (0.24)^y = 10^z ⇒ 0.24 = \(10^{\frac{z}{y}}\)

∴ \(\frac{2.4}{0.24}=\frac{10^{\frac{z}{x}}}{10^{\frac{z}{y}}} ⇒ 10 = 10^{\frac{z}{x}-\frac{z}{y}}\)

⇒ 1 = \(\frac{z}{x}-\frac{z}{y} = z\big(\frac{1}{x}-\frac{1}{y}\big).\)

⇒ \(\frac{1}{x}-\frac{1}{y}=\frac{1}{z} \,or\,\) \(\frac{1}{x}-\frac{1}{z}=\frac{1}{y}.\)

23000000

A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.

Then,

23000000 = 2.3 × 10⁷

Diameter of Earth = 12756000 m

A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.

Then,

Diameter of Earth = 12756000 m

= (1.2156 × 10⁷) m (in standard form)

(c) \(\frac94\)
\(x^{x\sqrt{x}} = (x\sqrt{x})^x\)

⇒ \(x^{x\sqrt{x}} = (x^{\frac32})^x\) ⇒ \(x^{x\sqrt{x}} =x^{\frac{3}{2}x}\)

⇒ \(x\sqrt{x} = \frac32x ⇒\sqrt{x} =\frac32⇒x=\frac94.\)