SChand CLASS 9 Chapter 6 Indices/Exponent Exercise 6(C)

  Exercise 6 C

Question 1

(i) 50×41+813

(ii) 3(64)4(125)2

(iii) (93×163/2)1/6

(iv) 3(16)3/4×(125)2

(v) (32)2/5÷(216)2/3

Sol :

(i) 50×41+813

=1×14+(2×2×2)13

=14+(23)1/3=14+(23×13)

=14+21

=2+14

214


(ii) 3(64)4(125)2

=[(64)4(125)2]1/3

=[(64)4×1/3]×[(125)2×1/3]

=[(64)4/3]×(125)2/3

=(43)4/3×(53)2/3

=43×(4/3)×53×(2/3)

=44×52

=144×52=14×4×4×4×5×5

=16400


(iii) (93×163/2)1/6

=[{13)2]3×(24)3/2]1/6

=[(3)6×(2)6]1/6

=36×1/6×26×1/6

=31×21=13×2

=23


(iv) 3(16)3/4×(125)2

=3(24)3/4×(53)2

=[24×(3/4)×53×(2)]1/3

=[23×56]1/3

[123×56]1/3

1(23)1/2×(56)1/3=121×52

=12×5×5=150


(v) (32)2/5÷(216)2/3

=(25)2/5÷(63)2/3

=(2)2/5×5+(6)3×(2/3)

=22÷62

=(26)2=(62)2=6×62×2

364

=9




Question 2

(i) (4125)2/3

(ii) 93/23×(5)0(181)1/2

(iii)  (14)23 (8)2/3 х 40+(316)1/2

(iv) 163/4+2(12)1×30

(v) (81)3/4(132)2/5+(8)1/3(12)2(2)0

Sol:

(i) (4125)2/3

1(265625)1/4+(25364)0

=(453)2/3÷1(4454)1/2 +[(52)1/2(43)1/3]0

(45)3×(2/3)×(45)4×1/4+1

=(45)2×(45)1+1

=(54)2×(54)1+1

=(54)21+1

=(54)+1

=54+1=114+1

=214 Ans


(ii) 93/23×(5)0(181)1/2

=(32)3/23×1(134)1/2

=32×3/231(34)1/2

=3331(3)2

=273(3)2=2739

=27-12

=15 Ans


(iii)  (14)23 (8)2/3 х 40+(316)1/2

=(122)23(23)2/3×1+[(3)2(4)2]1/2

=1(2)2(2)  3(23×2/3)× 1+  (34)2^(1/2)

124 3×22×1+(34)1

=243×4+43=1612+43

=4+113=513


(iv) 163/4+2(12)1×3

=(24)3/4+2×21×1=23+2×2×1

=2×2×2+2×2=8+4=12 Ans


(v) (81)3/4(132)2/5+(8)1/3(12)2(2)0

=(34)3/4125(2/5)+(23×1/3)×1(2)2×1

=(3)31(2)2+22×221

=27(2)1+2×22

=27-4+8

=35-4

=31 Ans



Question 3

(a) Evaluate :  x1/2×y1×z2/3 , when x = 9, y = 2, and z = 8.

(b) Evaluate as a fraction :
(278)2/3(14)2+(5)0

(c) Evaluate as a fraction :
14+(0.01)1/2(27)2/3

Sol: 

(a) x1/2×y1×z2/3
x=9
y=2
z=8

(9)1/2×(2)1×(8)2/3
=(32)1/2×12×(23)2/3
=32×1/2×12×23×2/3
=31×12×22=3×12×4
=3×2=6 Ans

(b) (278)2/3(14)2+(5)0
=(3323)2/3(122)2+1
33×2/323×2/3121×(2)+1
=3222124+1=9424+1
=9416+1=21416+1
=1234 Ans.

(c) 14+(0.01)1/2(27)2/3
= (122)1/2+(0.1)2×(1/2)(33)2/3
=122×1/2+(0.1)132
=12+1(0.1)9=12+1019
=10129=112 Ans

Question 4

(i) (64)5n/6(27)n/6(12)n/226n

(ii) 6(8)n+1+1623n21023n+17(8)n

(iii) 42n2n+12n342n+1

(iv) (5)2n+3(25)n+2((125)n+1]2/3

Sol: 

(i) (64)5n/6(27)n/6(12)n/226n

=(26)5n/6(39)n/6(22×3)n/226n

(2)6×5n6(3)3×(n6)(2)n2×2(3)n226n

=25n3n/22n3n/226n=(2)5n+6n+n(3)n2+n2

==212n30=(12)12n1.

=(2)12n


(ii) 6(8)n+1+1623n21023n+17(8)n

=6(23)n+1+1623n21023n+17(23)n

623n+3+1623n21023n+1723n

=623n23+1623n221023n2(72)3n

=23n(6×23+16×22)23n(10×27)

=6×8+16×14207

=48+413=5213=4


(iii) 42n2n+12n342n+1

=(22)2n2n+12(n3)(22)2n1
==24n2n+12n324n+2=24n+n+12n3+4n+2
=25n+125n1
=25n+15n+1
=22
=4


(iv) (5)2n+3(25)n+2((125)n+1]2/3

=53n+3(5)m+2[(53)n+3]2/3

=52n5352n545(3n+5)3/2

52n5352n5452n52

=52n(5354)52n52

=12562525

=50025

=-20 Ans



Question 5

If  49n+17n(343)n73m2n=3343  , prove that m = (m + 1)

Sol:

49n+17n(343)n73m2n3343(72)n+17n(73)n73m2n=33437n+27n73n73m24334373n(721)1673m=334373n×481673m=3343

=73n73m=3×16343×48=1343

=(7n)3(7m)3=173=(17)37n7m=177.7n=7m

7n+1=7m (comparing both sides) m=n+1 

Hence proved.



Question 6

If 2205=3a×5b×7c

(i) Find the numerical values of a, b and c

(ii) Hence evaluate 3a×5b×7c

Sol:
(i) 2205=3a×5b×7c

3220537355245749771

On comparing = a=2
b=1
c=2 
so  32×51×72=3a×5b×7c

(ii) 
3a×5b×7c=3a5b×7c=3251×72=95×4992459245 Ans. 



Question 7

If [b3c2b4c3]3+[b1cb2.c2]5=bx.cx , prove that x + y + 6 = 0

Sol:

[b3c2b4c3]3÷[b1cb2c2]5=bxcy
Now  R.H.S. =x+y+6=6+0+6=0
Hence proved.


Question 8

Prove that

(i) (x+y)1(x1+y1)=1xy

(ii) (x1+y1)1=xyx+y

(iii) a+b+ca1b1+b2c1+c1a1=abc

(iv) 11+xba+xca+11+xab+xcb+11+xbc+xac=1

(v) abxaxbbcxbxccax2xa=1


Sol:
(i) 
(x+y)1(x1+y1)=1xy L.H.S. =(x+y)1(x1+y2)=1x+y×(1x+1y)=1x+y×y+5xy

=1x+y×x+yxy1xy

Hence pruerd,
 L.H.S = R.H.S. 

(ii) (x1+y1)1=xyx+y

L.H.S 
(x1+y1)1=[1x+1y]1=[y+xxy]1=xyx+y R.H.S. 

(iii) a+b+ca1b1+b2c1+c1a1=abc

L.H.S
a+b+ca1b1+b1c1+c1a1=a+b+c1a1b+1b+1c+1c1a
=a+b+cc+a+babc(a+b+c)×abc(a+b+c)
=abc R.H.S (Hence proved)

(iv) 11+xba+xca+11+xab+xcb+11+xbc+xac=1
 
L.H.S
=1xaa+xbaxca+1xbb+xab+xcb+1xcc+xbc+xac

=1xa(xa+xb+xc)+1xb(xb+xa+xc)+1xc(x2+xa+xb)
=1(xa+xb+xc)[1xa+1xb+1xc]

1(xa+xb+x2)[xa+xb+xc1]=1
L.H.S =R.H.S PROVED

(v) abxaxbbcxbxccax2xa=1
xababxbcbcxcaca
=xabab+bcbc+caca
xabcbc+abac+bcababc
xDabc

=x0=1LHS=RHS

Hence, proved


Question 9

(i) (p1/3p1/3)(p2/3+1+p2/3)

(ii) (11+3)1/3(113)1/3

Sol: 

(i)(p1/3p1/3)(p2/3+1+p2/3)

if p1/3=x and p1/3=y
=(a2y2)(x2+xy+y2)
=x3y3=(p1/3)3(p1/3)3 =p1p1=p1p Ans.

(ii) (11+3)1/3(113)1/3
if 11=x
3=y
=(x+y)1/3(xy)1/3
[ from formula. ](x2y2)1/3
=[(11)2(3)2]1/3
=(113)1/3=(8)1/3
=(23)1/3=23/13
=22=2 Ans



Question 10

If 3x=5y=45z , prove that x=2yzyz

Sol: 

L.H.S 
Suppose 3x=5y=45z=a 
so, a1/x=3,a1/y=5,1a2=4545=3×3×532×5=45=a1z=a2x×a1/y=a1z=a2x+1y
In comparing
12=2x+1y=2x=1z1y 2x=yzyz 1x=yz2yz x=2yzyz R.H.S. 

Hence proved



Question 11

If x=abp1,y=abq1 and z=abr1 prove that xprxyrp.zpq=1.
Sol;
L.H.S.
x=abp1y=abq1z=abr1

So =xrryrpzpq
=(abp1)q˙r(ab21)rp(abr1)pq
=aqrb(2r)(p1)arpb(rp)(r1)ap2b(p2)(r1)
=ab
=1×1=1 R.H.S 
Hence proved

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