The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 3 Worksheet 2 Solutions of Exponents and Radicals offer comprehensive answers to textbook questions.
DAV Class 8 Maths Ch 3 WS 2 Solutions
Question 1.
Simplify:
(i) \(x^{1 / 2} \times x^{5 / 2}\)
Solution:
\(x^{1 / 2} \times x^{5 / 2}=x^{\frac{1}{2}+\frac{5}{2}}\)
= x3 [∵ am × an = am+n]
(ii) \(\frac{x^{6 / 5}}{x^{1 / 5}}\)
Solution:
\(\frac{x^{6 / 5}}{x^{1 / 5}}=x^{\frac{6}{5}-\frac{1}{5}}\) = x
[∵ am × an = am+n
(iii) (x7)0
Solution:
(x7)0 = x7 × 0 [∵ (xm)n = xmn]
= x0 = 1
(iv) \(5 x^{5 / 6} \times 6 x^{1 / 6}\)
Solution:
\(5 x^{5 / 6} \times 6 x^{1 / 6}\) = 5 × 6 × \(x^{\frac{5}{6}+\frac{1}{6}}\)
= 30 x6/6 = 30x.
(v) x-7/2 × 2x-1/2
Solution:
x-7/2 × 2x-1/2 = 1 × 2 . x-7/2 – 1/2
= 2 \(x^{\frac{-7-1}{2}}\)
= 2x-8/2 = 2x-4.
Question 2.
Find the value of
(i) (512)-2/9
Solution:
(512)-2/9 = (29)-2/9
= \(2^{9 \times \frac{-2}{9}}\) = 2-2
= \(\frac{1}{2^2}=\frac{1}{4}\)
(ii) [(216)2/3]1/2
Solution:
[(216)2/3]1/2 = [(63)2/3]1/2
= \(\left[6^{3 \times \frac{2}{3}}\right]^{1 / 2}=6^{2 \times \frac{1}{2}}\) = 6
(iii) 10 + 8-1/3
Solution:
10 + 8-1/3 = 10 + (23)-1/3
= 10 + 2-1 = 10 + \(\frac{1}{2}\)
= 10 × 2 = 20
(iv) (16)3/4
Solution:
(16)3/4 = (24)3/4
= 23 = 8
(v) 271/3 × 16-1/4
Solution:
271/3 × 16-1/4 = (33)-1/3 × (24)-1/4
= \(3^{3 \times \frac{-1}{3}} \times 2^{4 \times \frac{-1}{4}}\)
= 3-1 × 2-1
= \(\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)
(vi) \(\frac{1}{\left[\left(3^4\right)^{1 / 2}\right]^{-2}}\)
Solution:
\(\frac{1}{\left[\left(3^4\right)^{1 / 2}\right]^{-2}}=\frac{1}{\left[3^{4 \times 1 / 2}\right]^{-2}}\)
= \(\frac{1}{\left(3^2\right)^{-2}}=\frac{1}{3^{-4}}\)
= 34 = 81
(vii) \(\frac{27^{-2 / 3} \times 81^{5 / 4}}{\left(\frac{1}{3}\right)^{-3}}\)
Solution:
\(\frac{27^{-2 / 3} \times 81^{5 / 4}}{\left(\frac{1}{3}\right)^{-3}}=\frac{\left(3^3\right)^{-2 / 3} \times\left(3^4\right)^{5 / 4}}{(3)^3}\)
= \(\frac{3^{3 \times \frac{-2}{3}} \times 3^{4 \times \frac{5}{4}}}{27}=\frac{3^{-2} \times 3^5}{27}\)
= \(\frac{3^{5-2}}{27}=\frac{3^3}{27}=\frac{27}{27}\) = 1
(viii) 641/2 (641/2 + 1)
Solution:
641/2 (641/2 + 1) = \(8^{2 \times \frac{1}{2}}\left(8^{2 \times \frac{1}{2}}+1\right)\)
= 8 (8 + 1) = 8 × 9 = 72
(ix) \(\frac{36^{7 / 2}-36^{9 / 2}}{36^{5 / 2}}\)
Solution:
\(\frac{36^{7 / 2}-36^{9 / 2}}{36^{5 / 2}}=\frac{36^{7 / 2}}{36^{5 / 2}}-\frac{36^{9 / 2}}{36^{5 / 2}}\)
= 367/2 – 5/2 – 369/2 – 5/2
= \(36^{\frac{7-5}{2}}-36^{\frac{9-5}{2}}\)
= 362/2 – 364/2
= 36 – 362
= 36 – 1296 = – 1260
(x) 4 × 81-1/2 (811/2 + 813/2)
Solution:
4 × 81-1/2 (811/2 + 813/2)
= 4 × \(9^{2 \times \frac{-1}{2}}\left(9^{2 \times \frac{1}{2}}+9^{2 \times \frac{3}{2}}\right)\)
= 4 × 9-1 (9 + 93)
= 4 × (9-1+1 + 9-1+3)
= 4 × (90 + 92)
= 4 × (1 + 81)
= 4 × 82 = 328.
Question 3.
Evaluate:
(i) (0.04)3/2
Solution:
(0.04)3/2 = [(0.2)2]3/2
= \((0.2)^{2 \times \frac{3}{2}}\)
= (0.2)3 = 0.008
(ii) (6.25)3/2
Solution:
(6.25)3/2 = [(2.5)2]3/2
= \((2.5)^{2 \times \frac{3}{2}}\)
= (2.5)3 = 15.625
(iii) (0.03125)-2/5
Solution:
(0.03125)-2/5 = [(2.5)5]-2/5
= \((2.5)^{5 \times \frac{-2}{5}}\)
= (2.5)-2
= \(\frac{1}{(2.5)^2}=\frac{1}{6.25}\)
(iv) (0.008)2/3
Solution:
(0.008)2/3 = [(0.2)3]2/3
= (0.2)3 × \(\frac{2}{3}\)
= (0.2)2 = 0.04
Question 4.
Evaluate:
(i) (62 + 82)1/2
Solution:
(62 + 82)1/2 = (36 + 64)1/2
= (100)1/2
= \(10^{2 \times \frac{1}{2}}\) = 10
(ii) \(\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^3\right]^{1 / 4}\)
Solution:
\(\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^3\right]^{1 / 4}=\left[5\left(2^{3 \times \frac{1}{3}}+3^{3 \times \frac{1}{3}}\right)^3\right]^{1 / 4}\)
= [5(2 + 3)3]1/4
= [5 × 53]1/4
= \(5^{4 \times \frac{1}{4}}\) = 5
(iii) (172 – 82)1/2
Solution:
(172 – 82)1/2 = (289 – 64)1/2
= (225)1/2
= 152 × 1/2 = 15
(iv) (13 + 23 + 33)-5/2
Solution:
(13 + 23 + 33)-5/2 = (1 + 8 + 27)-5/2
= (36)-5/2
= \(6^{2 \times\left(-\frac{5}{2}\right)}\) = 6-5
= \(\frac{1}{6^5}=\frac{1}{7776}\)
Question 5.
Simplify and express the answers with positive indices:
(i) 2x1/6 × 2. x-7/6
Solution:
2x1/6 × 2. x-7/6 = 2 × 2 × \(x^{\frac{1}{6}-\frac{7}{6}}\)
= \(4 x^{\frac{1-7}{6}}\)
= 4x-1 = \(\frac{4}{x}\)
(ii) \(\left[\sqrt[4]{\left(\frac{1}{x}\right)^{-12}}\right]^{-2 / 3}\)
Solution:
\(\left[4 \sqrt{\left(\frac{1}{x}\right)^{-12}}\right]^{-2 / 3}=\left[\left(\frac{1}{x}\right)^{-12 \times \frac{1}{4}}\right]^{-2 / 3}\)
= \(\left[\left(\frac{1}{x}\right)^{-3}\right]^{-2 / 3}=\left(\frac{1}{x}\right)^{-3 \times-2 / 3}\)
= \(\left(\frac{1}{x}\right)^2=\frac{1}{x^2}\)
(iii) a4/7 + a10/7
Solution:
a4/7 + a10/7 = \(a^{\frac{4}{7}-\frac{10}{7}}\)
= \(a^{\frac{-6}{7}}=\frac{1}{a^{6 / 7}}\)
Question 6.
Verify that [(729)-5/3]-1/2 = (729)-5/3 × (-1/2)
Solution:
L.H.S. = [(729)-5/3]-1/2
= 9[(93)-5/3]-1/2
= [93 × -5/3]-1/2
= [9-5]-1/2 = 95/2
= (32)5/2
= \(3^{2 \times \frac{5}{2}}\)
= 35 = 243
R.H.S. = (729)-5/3 × (-1/2)
= (729)5/6
= (36)5/6
= 36 × 5/6
= 35 = 243
LHS = RHS
Hence, verified.
Question 7.
Solve the given exponential equations:
(i) (√6)x-2 = 1
Solution:
(√6)x-2 = 1 = \(\left[6^{\frac{1}{2}}\right]^{x-2}\)
= 1 = \(6^{\frac{x-2}{2}}\) = 60
Equating the power on both sides,
\(\frac{x-2}{2}\) = 0
⇒ x – 2 = 0
∴ x = 2
(ii) 34x = \(\frac{1}{81}\)
Solution:
34x = \(\frac{1}{81}\)
⇒ 34x = \(\frac{1}{3^4}\)
⇒ 34x = 3– 4
Equating the power on both sides,
4x = – 4
x = \(\frac{-4}{4}\)
∴ x = – 1
(iii) (√2)x = 28
⇒ (21/2)x = 23
⇒ 2x/2 = 23
Equating the power on both sides,
\(\frac{x}{2}\) = 3
∴ x = 2 × 3 = 6
(iv) 22x – 1 = 42x – 1
Solution:
22x – 1 = 42x – 1
⇒ 22x – 1 = [(2)2]2x – 1
⇒ 22x – 1 = 24x – 2
Equating the power on both sides,
2x – 1 = 4x – 2
2x – 4x = – 2 + 1
⇒ – 2x = – 1
⇒ 2x = 1
∴ x = \(\frac{1}{2}\).
DAV Class 8 Maths Chapter 3 Value Based Questions
Question 1.
As part of ‘Earth Day’ celebrations an interclass poster making competition was arranged between nine sections of Class-VIII. Each section was provided a sheet measuring 75 cm × 40 cm for making the poster.
(i) Find the total area of sheet provided to the students. Express the area in exponential form.
(ii) Write any two ways by which you can save earth.
Solution:
(i) Area of one sheet = length × breadth
= 75 cm × 40 cm = 3000 cm2
Total area of 9 sheets = 9 × 3000 = 27000 cm2 = (30)3 cm2
(ii) By keeping environment clean, and planting more and more trees, we can save our earth.