WBBSE Solutions For Class 9 Maths Algebra Chapter 1 Laws Of Indices

Algebra Chapter 1 Laws Of Indices

If a certain real number x is multiplied m times in succession (where m is a positive integer) then the continued product so obtained is called the mth power of x and is written by x^m.

Then x = x x x x x x …………. to m factors.

Here x is called the base of x^m and m is called the index or exponent of x^m.

Laws of indices:

If a, b are two non zero real numbers and m, n are positive integers them

1. a^m, aⁿ = a^m+n (This rule is called the fundamental law of index.)
2. a^m ÷ aⁿ = a^m-n
   
3. (a^m)ⁿ = a^mn
4. (ab)^m = a^m. b^m
5. \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
6. a⁰ = 1 (a ≠ 0).
7. If a, m, n are three real numbers and a^m = aⁿ (a ≠ 0, 1, -1), thrn m = n.
8. If a, b, and m are three real numbers and a^m = b^m, then either a = b or m = 0
9. \(\sqrt[q]{a^p}=a^{\frac{p}{q}}\) (p, q are positive integers)
10. \(a^{-m}=a^{\frac{1}{m}} \quad(a \neq 0)\)

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Algebra Chapter 1 Laws Of Indices True Or False

Example 1. In x^m, x is called index and m is called the basc.

Solution: The statement is False.

Example 2. \(\left(\frac{a}{b}\right)^2=\left(\frac{b}{a}\right)^{-2}\)

Solution: The statement is True.

Example 3. x⁰ = 1 for any real number x.

Solution: The statement is False.

Example 4. If a^x = a^y then x = y for any real number a.

Solution: The statement is False.

Example 5. \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\)

Solution: The statement is True.

Example 6. If a^x = k then a = k^-x

Solution: The statement is False.

Example 7. If 3^x = 2^x then x = 0

Solution: The statement is True.

Example 8. \(a^{-\frac{p}{q}}=\sqrt[q]{a^p}\) (p, q are positive integers a ≠ 0)

Solution: The statement is True.

Example 9. \((-27)^{\frac{1}{3}}=-3\)

Solution: The statement is True.

Example 10. If m, n, and p are positive integers, then a^m. aⁿ. a^p = a^m+n+p.

Solution: The statement is True.

Algebra Chapter 1 Laws Of Indices Fill In The Blanks

Example 1. Value of \((81)^{\frac{3}{4}}\) is ________

Solution: 27.

Example 2. Value of p⁰ (p ≠ 0) is ________

Solution: 1

Example 3. \(\sqrt[3]{\left(\frac{1}{64}\right)^2}\) = ___________

Solution: \(\frac{1}{2}\)

Example 4. If (27)^x = (81)^x then x: y

Solution: 4:3.

Example 5. If x = 5 and y = 3 then \((x+y)^{\frac{x}{y}}\) is ___________

Solution: 32.

Example 6. If a ≠ b ≠ 0, and a^x = b^x then x ________

Solution: 0.

Example 7. If 4^x = 8³ then x = ________

Solution: \(\frac{9}{2}\)

Example 8. \(2^{\frac{1}{2}}\times 2^{-\frac{1}{2}} \times(64)^{\frac{1}{6}}\) = ________

Solution: 1

Example 9. √√2 = ________

Solution: \(2^{\frac{1}{4}}\)

Example 10. \(x^{a-b} \times x^{b-c} \times x^{c-a}\) = _______

Solution: 1.

Algebra Chapter 1 Laws Of Indices Short Answer Type Questions

Example 1. If (5⁵+0.01)²– (5²0-0.01)² = 5^x then find the value of x.

Solution: 4.5⁵.0.01 = 5^x [(a + b)² – (a – b)² = 4.ab]

⇒ or, \(\frac{4 \times 5}{100}=5^x\)

⇒ or, 5^5-2 = 5^x

⇒ or, x=3

The value of x is 3

Example 2. If 3 x 27^x = 9^x+4 then find the value of x.

Solution: 3 x (3³)^x = (3²)^x+4

or, 3^1+3x = 3^2x+8

∴ 1+3x= 2x + 8

or, x = 7.

The value of x is 7.

Example 3. Which is greater \(3^{3^3} \text { or, }\left(3^3\right)^3 \text { ? }\)

Solution: \(3^{3^3}=3^{27}, \quad\left(3^3\right)^3=3^9, 27>9\)

∴ \(3^{3^3}>\left(3^3\right)^3\)

Example 4. \((\sqrt[5]{8})^{\frac{5}{2}} \times(16)^{-\frac{3}{8}}\) = ?

Solution: = \(\left(8^{\frac{1}{5}}\right)^{\frac{5}{2}} \times\left(2^4\right)^{-\frac{3}{8}}\)

= \(8^{\frac{1}{5} \times \frac{8}{2}} \times 2^{4 \times\left(-\frac{3}{8}\right)}\)

= \(\left(2^3\right)^{\frac{1}{2}} \times 2^{-\frac{3}{2}}=2^{\frac{2}{2}+\left(-\frac{3}{2}\right)}=2^0=1\)

\((\sqrt[5]{8})^{\frac{5}{2}} \times(16)^{-\frac{3}{8}}\) = \(\left(2^3\right)^{\frac{1}{2}} \times 2^{-\frac{3}{2}}=2^{\frac{2}{2}+\left(-\frac{3}{2}\right)}=2^0=1\)

Example 5. Simplify \(\sqrt[5]{x^8 \cdot \sqrt{x^6 \cdot \sqrt{x^4}}}\)

Solution: \(\sqrt[5]{x^8 \sqrt{x^6 x^{-\frac{4^2}{2}}}}\)

= \(\sqrt[5]{x^8 \sqrt{x^{6-2}}}=\sqrt[5]{x^8 \sqrt{x^4}}\)

= \(\sqrt[5]{x^8 x^{\frac{24}{2}}}\)

= \(\sqrt[5]{x^{8+2}}=x^{\frac{10}{5}}=x^2\)

\(\sqrt[5]{x^8 \cdot \sqrt{x^6 \cdot \sqrt{x^4}}}\) = \(\sqrt[5]{x^{8+2}}=x^{\frac{10}{5}}=x^2\)

Example 6. Simplify \(\left\{(81)^{-\frac{3}{4}} \times \frac{16^{\frac{1}{4}}}{6^{-2}} \times\left(\frac{1}{27}\right)^{\frac{4}{3}}\right\}^{\frac{1}{3}}\)

Solution: =\(\left\{3 \times\left(-\frac{3}{4}\right) \times \frac{2^{4 \times \frac{1}{4}}}{3^{-2} \times 2^{-2}} \times\left(3^{-3}\right)^{-\frac{4}{3}}\right\}^{\frac{1}{3}}\)

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

= \(\left\{3^{-3+4+2} \times 2^{1+2}\right\}^{\frac{1}{3}}\)

= \(\left(3^3 \times 2^3\right)^{\frac{1}{3}}=6^{3 \times \frac{1}{3}}=6\)

Example 7. If \(x^{p^q}=\left(x^p\right)^q\), find p in terms of q, (x ≠ 0, 1, −1)

Solution: \(x^{p^q}=x^{p q}\)

⇒ or, \(xp^q=p q \quad(x \neq 0,1,-1)\)

⇒ or, \(\frac{p^q}{p^1}=q\)

⇒ or, \(p^{q-1}=q\)

∴ \(p=q-1 \sqrt{q}\)

Example 8. Arrange in the ascending order of magnitude, 2⁶³, 3⁴⁵, 5²⁷, 6¹⁸

Solution: 2⁶³ = (2⁷)⁹ = (128)⁹

⇒ 3⁴⁵ = (3⁵)⁹ = (243)⁹

⇒ 5²⁷ = (5³)⁹ = 125⁹

⇒ 6¹⁸ = (6²)⁹ = (36)⁹

[Note: You have to find out the H.C.F. of 63, 45, 27, 18 which is 9]

⇒ Since 36125 < 128 < 243

⇒ Hence 6¹⁸ < 5²⁷ < 2⁶³ < 3⁴⁵

Example 9. If \(x^{x \cdot \sqrt{x}}=(x \sqrt{x})^x\), find the value of x.

Solution: \((x \sqrt{x})^x=(x \sqrt{x})^x\)

⇒ or, \(x^{\sqrt{x}}=x \sqrt{x}\)

⇒ Now, \(x \sqrt{x}=x^{1+\frac{1}{2}}=x^{\frac{3}{2}}\)

∴ \(\sqrt{x}=\frac{3}{2}, \quad x=\frac{9}{4}\)

The value of x  \(\sqrt{x}=\frac{3}{2}, \quad x=\frac{9}{4}\)