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(i) Given 13²

13² 

= 13 × 13 

= 169

(ii) Given 7³

7³ 

= 7 × 7 × 7 

= 343

(iii) Given 3⁴

3⁴ 

= 3 × 3 × 3 × 3

= 81

(c) \(\frac{1}{\sqrt[3]{c^4}}\)

a = \(b^{\frac{2}{3}}\) and b = c^-2

∴ a = (c^-2)^{\({\frac{2}{3}}\)} = c^\(-\frac43\) = \(\frac{1}{c^{\frac43}}\) = \(\frac{1}{\sqrt[3]{c^4}}\)

(i)  Given 3² × 10⁴ 

3² × 10⁴ 

= 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

(ii) Given 2⁴ × 3² 

2⁴ × 3² 

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(iii) Given 5² × 3⁴ 

5² × 3⁴ 

= 5 × 5 × 3 × 3 × 3 × 3

= 25 × 81

= 2025

(i) Given (-7)²

We know that (-a) ^{even number}= positive number

(-a) ^{odd number} = negative number

We have, (-7)² 

= (-7) × (-7)

= 49

(ii) Given (-3)⁴

We know that (-a) ^{even number}= positive number

(-a) ^{odd number} = negative number

We have, (-3)⁴ 

= (-3) × (-3) × (-3) × (-3)

= 81

(iii) Given (-5)⁵

We know that (-a) ^{even number}= positive number

(-a) ^{odd number} = negative number

We have, (-5)⁵ 

= (-5) × (-5) × (-5) × (-5) × (-5)

= -3125

\(\frac{1}{({625)}^{\frac{1}{4}}}\) = \(\frac{1}{({5^4)}^{\frac{1}{4}}}\) = \(\frac{1}{5}\)

= 0.2

(4/3)^-3× (4/3)^-2

We have,

= (4/3) ^{(-3+ (-2))} … [{(a/b) ^m × (a/b) ⁿ} = (a/b) ^m-n

= (4/3) ^(-3-2)

= (4/3)^-5

= (3/4)⁵

= (3⁵/4⁵)

= 243/1024

i. 3^1+2+3+4+5+6 = 3²¹ 

[a^m x aⁿ = a^m+n] 

ii. 2² × 3³ × 2⁴ × 3⁵ × 3⁶ [a^m x aⁿ = a^m+n] 

= 2²⁺⁴ × 3³⁺⁵⁺⁶ 

= 2⁶ × 3¹⁴

(-7/8)^-3× (-7/8)²

We have,

= (-7/8) ^{(-3+ 2)} … [{(a/b) ^m × (a/b) ⁿ} = (a/b) ^m-n

= (-7/8) ^(-1)

= (-8/7)

(c) \(6\frac14\)

Given exp. = \(\frac{5.(5^2)^{n+1}+5^2.5^{2n+1}}{5^2.5^{2n}-21\times5(5^2)^{n-1}} \)

= \(\frac{5^{2n+3}+5^{2n+1}}{5^{2n+2}-21\times5^{2n-1}}\) = \(\frac{5^{2n+1}(5^2+1)}{5^{2n-1}(5^3-21)}\)

= \(\frac{5^{(2n+1)-(2n-1)}\times26}{(125-21)}\) = \(\frac{5^2\times26}{104}=\frac{25}{4}=6\frac14.\)

(i) Given 3 × 10²

3 × 10² 

= 3 × 10 × 10

= 3 × 100

= 300

(ii) Given 2² × 5³

2² × 5³ 

= 2 × 2 × 5 × 5 × 5

= 4 × 125

= 500

(iii) Given 3³ × 5²

3³ × 5² 

= 3 × 3 × 3 × 5 × 5

= 27 × 25

= 675

(4/9)⁶× (4/9)^-4

We have,

= (4/9) ^{(6+ (-4))} … [{(a/b) ^m × (a/b) ⁿ} = (a/b) ^m-n

= (4/9) ^(6-4)

= (4/9)²

= (4²/9²)

= (16/81)

(d) 1

 8^x–2 x \(\big(\frac12\big)^{4-3x}\) = (0.0625)^x 

(2³)^{x – 2} x (2^–1)^4–3x = \(\big(\frac{625}{10000}\big)^x\) 

⇒ 2^3x-6 x 2^-4+3x =  \(\big(\frac{1}{16}\big)^x\) = (2^-4)^x = 2^-4x

⇒ 2^3x-6-4+3x = 2^-4x

⇒ 2^{6x – 10} = 2^{– 4x} 

⇒ 6x – 10 = – 4x 

⇒ 10x = 10 ⇒ x = 1.

(a) 4

\(5\sqrt5\times5^3÷5^{-\frac32} = 5^{a+2}\)

⇒ \(5^{\frac32}\times5^3÷5^{-\frac32} = 5^{a+2}\)

⇒ \(5^{\frac32+3-(-\frac32)} = 5^{a+2}\)

⇒ 5⁶ = 5^a+2  ⇒ a + 2 = 6 ⇒ a = 4

(b) 12

2^x - 2^x-1 = 4 ⇒ 2^x -\(\frac{2^x}{2}\) = 4 

⇒ 2^x\(\big(1-\frac12\big)\) = 4 ⇒ 2^x x \(\frac12\) = 4

⇒ 2^x = 8 ⇒ 2^x = 2³ ⇒ x = 3

∴ 2^x + 2^x-1 = 2³ + 2² = 8 + 4 = 12.

(a)  \(3^{2^{2^{2}}}\) =  \(3^{2^{4}}\) = 3¹⁶;

\(\{(3^2)^2\}^2 = 3^8 = 6561;\)

3² × 3² × 3² = 3^{2 + 2 + 2} = 3⁶ = 729 ;

∴ 3¹⁶ > 3⁸ > 3222 > 3⁶

{(-3/4)³-(-5/2)³} × 4²

We have,

= {(-3³/4³)-(5³/2³)} × 16

= {(-27/64)-(-125/8)} × 16

First we find the difference of {(-27/64)-(125/8)}

LCM of 64 and 8 is 64

= (-27×1) / (64×1) = (-27/64)

= (125×8) / (8 ×8) = (1000/64)

= (1000- (-27))/64

= (1000+27)/64

= (973/64)

= (973/64) × 16

= (973/4)

(2/3)² × (-3/5)³ × (7/2)²

We have,

(2²/3²) = (2×2)/ (3×3) = (4/9)

(-3/5)³= (-3×-3×-3) / (5×5×5) = (-27/125)

(7²/2²) = (7×7)/ (2×2) = (49/4)

Then,

= (4/9) × (-27/125) × (49/4)

= (4×-27×49) / (9× 125×4)

On simplifying,

= (1×-3×49) / (1×125×1)

= (-147/125)

0.15 = \(\frac{1.5}{10}=\frac{3}{2\times10}=\frac{10^n}{10^m\times10}=\frac{10^n}{10^{m+1}}=10^{n-(m+1)}=10^{n-m-1}\)

(c) 3125

6^x - 6^x-3 = 7740 ⇒ 6^x - \(\frac{6^x}{6^3}\) = 7740

⇒ \(6^x\bigg(1-\frac{1}{216}\bigg)=7740 ⇒ 6^x\times \frac{215}{216}=7740\)

⇒ \(6^x = \frac{7740\times216}{215}\) = 36 x 216 = 6⁵ ⇒ x = 5

∴ \(x^x = 5^5\) = 3125.

(d) 1

\(\bigg(2^{\frac14}-1\bigg)\bigg(2^{\frac34}+2^{\frac12}+2^{\frac14}+1\bigg)\)

Let \(2^{\frac14}\) = a . Then, 

Given exp. = (a – 1) (a³ + a² + a + 1 ) 

= (a – 1) (a² (a + 1) + 1(a + 1)) 

= (a – 1) (a + 1) (a² + 1) = (a² – 1) (a² + 1) 

= a⁴ – 1

∴ Required value = \(\big(2^{\frac14}\big)^4\) -1 = 2 - 1 = 1.

(3/2)⁴ × (1/5)²

We have,

(3⁴/2⁴) = (3×3×3×3)/ (2×2×2×2) = (81/16)

(1²/5²) = (1×1)/ (5×5) = (1/25)

Then,

= (81/16) × (1/25)

= (81×1) / (16/25)

= (81/400)

ac = b²

⇒ 2^x .8^z = (4^y)² ⇒ 2^x .(2³)^z = ((2²)^y)² ⇒ 2^x . 2^3z = 2^4y 

⇒ 2^x+3z = 2^4y ⇒ x + 3z = 4y.

(b) \(\frac{17}{8}\), \(\frac{15}{8}\)

3^{x + y} = 81 ⇒ 3^{x + y} = 3⁴ = x + y = 4     ... (i) 

81^{x – y} = 3 ⇒ (3⁴)^{x – y} = 3¹

⇒ 4x – 4y = 1                     ... (ii) 

Eqn (i) × 4 + Eqn (ii) gives 

4x + 4y + 4x – 4y = 16 + 1 

⇒ 8x = 17 ⇒ x = \(\frac{17}{8}\) 

Putting x = \(\frac{17}{8}\) in (i), we get \(\frac{17}{8}\) + y = 4

⇒ y = 4 - \(\frac{17}{8}\) = \(\frac{15}{8}\)

∴ x = \(\frac{17}{8}\), y = \(\frac{15}{8}\).

1 / (100/9)^3/2

= (10/3) ^{-3/2 × 2}

\(= \frac{3}{10}\)\(\times\frac{3}{10}\)\(\times\frac{3}{10}\)

1 / (8)^{-1/2 × 1/3}

= 2 ^-1/2

= ^{\(\frac{1}{\sqrt{2}}\)}

(d) 8

\(\sqrt{3^n}\) = 81 ⇒ 3^n/2 = 3⁴ ⇒ \(\frac{n}2\) = 4 ⇒ n = 8

8 ^{x + 1} – 64 

= 8 ^{x + 1} = 8² 

On equating powers, we get 

x + 1 = 2 

x = 1 

= 3 ^{2x + 1} 

= 3³ = 27

Given, \(\bigg(3^{\frac12}\bigg)^5 \times(3^2)^2 =3^n\times3\times3^{\frac12}\)⇒ \(3^{\frac52}\times3^4=3^{n+1+\frac12}\)

⇒ \(3^{\frac52+4}=3^{n+\frac32}\) ⇒ \(3^{\frac{13}{2}}=3^{n+\frac32}\) ⇒ n+\(\frac32=\frac{13}{2}\) ⇒  n = \(\frac{13}{2}-\frac{3}{2} = \frac{10}2 = 5.\) 

(0.04)^–1.5 = (0.04)^\(-\frac32\) = \(\frac{1}{(0.04)^{\frac32}}=\frac{1}{\sqrt{(0.04)3}}\) 

= \(\frac{1}{(0.2)^3}=\frac{1}{0.008}=\frac{1000}{8}=125.\)

(x ^-1 + y ^-1) ^-1

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

= ^{\((\frac{x+y}{xy})^{-1}\)}

^{= \((\frac{xy}{x+y})\)}

Given, x^z = y² ⇒ (10^0.48)^z ⇒ (10^0.70)²

⇒ 10^0.48z = 10^1.40 ⇒ 0.48z = 1.40 ⇒ z = \(\frac{140}{48} = \) 2.9 (approx)

Given exp. = (10²)^\(\frac12\) x \((\frac{1}{1000})^\frac13-(\frac{16}{1000})^\frac14\times3^0+(\frac54)^{-1}\)

= 10^{2 x \(\frac12\)} x \((\frac{1}{10})^{3\times\frac13} - (\frac{2}{10})^{4\times\frac14}\times3^0+\frac45\)

= 10 x \(\frac{1}{10}-\frac15+\frac45\) = 10 × 0.1 – 0.2 + 0.8 = 1 + 0.6 = 1.6.

\(\{(\cfrac{5}{6})^{\cfrac{1}{5}}\}^{-\cfrac{1}{6}}\)

= 1 / {(5/6) ^1/5} ^1/5 

= (5/6) ^-1/30 

= (6/5) ^1/30

940000000000

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,

940000000000 = 9.4 × 10¹¹

x ^1/7 / x^1/8 

= (x) ^{1/7 – 1/8} 

= (x) ^1/56

^{\(= \sqrt[56]{x}\)}

As 64 can be written as 64 = 2×2×2×2×2×2

so

64 = 2⁶ \(\sqrt{64}\)

= \(\sqrt(2^6)\) 

=(2⁶) ^1/2

= 2³ = \(8\sqrt[3]{64}\)

= (2⁶)^1/3 

= \(2^2\) = \(4\frac{\sqrt{64}}{\sqrt[3]{64}}\)

= \(\frac{8}4\) = 2

Given exp. = \(\frac{6^{\frac23}\times6^{\frac73}}{6^{\frac63}}=6^{\frac23+\frac73-\frac63}=6^{\frac33}=6^1=6.\)

9 ^{x + 2} = 240 + 9^x 

9x × 9² = 240 + 9^x 

Let 9^x = y 

81y = 240 + y 

80y = 240

y = \(\frac{240}{80}\)

9^x = 3 

3^2x = 3 

2x = 1 

x = \(\frac{1}{2}\)

= 0.5

(-4)³

We know that the reciprocal of (a/b) ^m is (b/a) ^m

Then,

Reciprocal of (-4)³ is (-1/4)³

(6)⁷

We know that the reciprocal of (a/b) ^m is (b/a) ^m

Then,

Reciprocal of (6)⁷ is (1/6)⁷

 (1/4)^-4

We know that,

= (1/4)^-4 = (4/1)⁴ … [∵ (a/b)^-n = (b/a) ⁿ]

= (4⁴/1⁴)

= (256/1)

= 256

82934000000

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,

82934000000 = 8.2934 × 10¹⁰

0.0000463 in standard form is written as: 

0.0000463 = 0.463 × 10^-4 

= 4.63 × 10^-5

\(\sqrt{x}\times\sqrt[3]{x}\)

= x^1/2 × x^1/3 

= x^5/6

(-3/4)^-3

We know that,

= (-3/4)^-3 = (-4/3)³ … [∵ (a/b)^-n = (b/a) ⁿ]

= (-4³/3³)

= (-64/27)

(3/8)⁴

We know that the reciprocal of (a/b) ^m is (b/a) ^m

Then,

Reciprocal of (3/8)⁴ is (8/3)⁴

6428000

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,

6428000 = 6.428 × 10⁶

x – y ^{x – y} 

= 2 – (-2)^{(2 + 2)} 

= 2 – 16 

= - 14

(-3)^-1× (1/3)^-1

We know that,

= (-3)^-1= (-1/3)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (1/3)^-1 = (3/1)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (-1/3) × (3/1)

= (-1×3)/ (3×1)

= (-3/3)

= -1

(–2/3)^-1

We have:

(-2/3)^-1 = (-2/3)^-1

= (3/-2)¹ … [∵ (a/b)^-n = (b/a) ⁿ]

= (-3/2)