RD Sharma (2017) Solutions for Class 9 Math Chapter 2 Exponents Of Real Numbers are provided here with simple step-by-step explanations. These solutions for Exponents Of Real Numbers are extremely popular among class 9 students for Math Exponents Of Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma (2017) Book of class 9 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma (2017) Solutions. All RD Sharma (2017) Solutions for class 9 Math are prepared by experts and are 100% accurate.

Page No 2.12:

Question 1:

Simplify the following:

(i) 3a4b310×5a2b23

(ii) 2x-2y33

(iii) 4×1076×10-58×104

(iv) 4ab2-5ab310a2b2

(v) x2y2a2b3n

(vi) a3n-96a2n-4

Answer:

(i)
3a4b310×5a2b23=3×a40×b30×5×a6×b6=15×a40×a6×b30×b6=15×a40+6×b30+6              am×an=am+n=15a46b36

(ii)
2x-2y33=23×x-23×y33=8×x-6×y9=8x-6y9

(iii)
4×1076×10-58×104=4×107×6×10-58×104=24×107+-58×104=24×1028×104
=24×102×10-48=3×102+-4=3×10-2=3100

(iv)
4ab2-5ab310a2b2=4×a×b2×-5×a×b310a2b2=-20×a1×a1×b2×b310a2b2
=-20×a1+1×b2+310a2b2=-2×a2×b5×a-2×b-2=-2×a2+-2×b5+-2=-2×a0×b3=-2b3

(v)
x2y2a2b3n=x2ny2na2nb3n=x2ny2na2nb3n

(vi)
a3n-96a2n-4=a63n-9a2n-4=a18n-54a2n-4
=a18n-54×a-2n-4=a18n-54×a-2n+4=a18n-54-2n+4=a16n-50

Page No 2.12:

Question 2:

If a=3 and b=-2, find the values of:
(i) aa+bb
(ii) ab+ba
(iii) a+bab

Answer:

(i) aa+bb
Here, a=3 and b=-2.
Put the values in the expression aa+bb.
33+-2-2=27+1-22=27+14=108+14=1094

(ii) ab+ba
Here, a=3 and b=-2.
Put the values in the expression ab+ba.
3-2+-23=132+-8=19-8=1-729=-719

(iii) a+bab
Here, a=3 and b=-2.
Put the values in the expression a+bab.
3+-23-2=1-6=1

Page No 2.12:

Question 3:

Prove that:

(i) xaxba2+ab+b2×xbxcb2+bc+c2×xcxac2+ca+a2=1

(ii) xax-ba2-ab+b2×xbx-cb2-bc+c2×xcx-ac2-ca+a2=1

(iii) xaxbc×xbxca×xcxab=1

Answer:

(i) xaxba2+ab+b2×xbxcb2+bc+c2×xcxac2+ca+a2=1

Consider the left hand side:
xaxba2+ab+b2×xbxcb2+bc+c2×xcxac2+ca+a2=xaa2+ab+b2xba2+ab+b2×xbb2+bc+c2xcb2+bc+c2×xcc2+ca+a2xac2+ca+a2=xaa2+ab+b2-ba2+ab+b2×xbb2+bc+c2-cb2+bc+c2×xcc2+ca+a2-ac2+ca+a2=xa-ba2+ab+b2×xb-cb2+bc+c2×xc-ac2+ca+a2=xa3-b3×xb3-c3×xc3-a3=xa3-b3+b3-c3+c3-a3=x0=1
Left hand side is equal to right hand side.
Hence proved.

(ii) xax-ba2-ab+b2×xbx-cb2-bc+c2×xcx-ac2-ca+a2=1
Consider the left hand side:
xax-ba2-ab+b2×xbx-cb2-bc+c2×xcx-ac2-ca+a2=xa+ba2-ab+b2×xb+cb2-bc+c2×xc+ac2-ca+a2=xa3+b3×xb3+c3×xc3+a3=x2a3+b3+c3
Left hand side is not equal to one.
Disclaimer: The question given in the book is not correct, the left hand side can not be equal to the right hand side.

(iii) xaxbc×xbxca×xcxab=1
Consider the left hand side:
=xacxbc×xbaxca×xcbxab=xacxbc×xbaxca×xcbxab=xac×xba×xcbxbc×xca×xab=xac+ba+cbxbc+ca+ab=1
Left hand side is equal to right hand side.
Hence proved.

Page No 2.12:

Question 4:

Prove that:

(i) 11+xa-b+11+xb-a=1

(ii) 11+xb-a+xc-a+11+xa-b+xc-b+11+xb-c+xa-c=1

Answer:

(i) Consider the left hand side:
11+xa-b+11+xb-a=11+xaxb+11+xbxa=1xb+xaxb+1xa+xbxa=xbxb+xa+xaxa+xb=xb+xaxb+xa=1
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
11+xb-a+xc-a+11+xa-b+xc-b+11+xb-c+xa-c=11+xbxa+xcxa+11+xaxb+xcxb+11+xbxc+xaxc=1xa+xb+xcxa+1xb+xa+xcxb+1xc+xb+xcxc=xaxa+xb+xc+xbxb+xa+xc+xcxc+xb+xc=xa+xb+xcxa+xb+xc=1
Therefore left hand side is equal to the right hand side. Hence proved.

Page No 2.12:

Question 5:

Prove that:

(i) a+b+ca-1b-1+b-1c-1+c-1a-1=abc

(ii) a-1+b-1-1=aba+b

Answer:

(i) Consider the left hand side:
a+b+ca-1b-1+b-1c-1+c-1a-1=a+b+c1ab+1bc+1ca=a+b+cc+a+babc=a+b+c×abca+b+c=abc
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
a-1+b-1-1=1a-1+b-1=11a+1b=1b+aab=aba+b
Therefore left hand side is equal to the right hand side. Hence proved.

Page No 2.12:

Question 6:

If abc = 1, show that 11+a+b-1+11+b+c-1+11+c+a-1=1.

Answer:

Consider the left hand side:
11+a+b-1+11+b+c-1+11+c+a-1=11+a+1b+11+b+1c+11+c+1a=1b+ab+1b+11+b+ab+11+1ab+1a                 abc=1=bb+ab+1+11+b+ab+abab+1+b=b+1+abb+ab+1=1
Hence proved.

Page No 2.12:

Question 7:

Simplify the following:

(i) 3n×9n+13n-1×9n-1

(ii) 5×25n+1-25×52n5×52n+3-25n+1

(iii) 5n+3-6×5n+19×5x-22×5n

(iv) 68n+1+1623n-21023n+1-78n

Answer:

(i)
3n×9n+13n-1×9n-1=3n×32n+13n-1×32n-1=3n×32n+23n-1×32n-2
=3n+2n+23n-1+2n-2=33n+233n-3=33n+2-3n+3=35=243

(ii)
5×25n+1-25×52n5×52n+3-25n+1=5×52n+1-52×52n5×52n+3-52n+1=5×52n+2-52×52n5×52n+3-52n+2=51+2n+2-52+2n51+2n+3-52n+2
=52n+3-52+2n52n+4-52n+2=52+2n5-152n+252-1=424=16

(iii)
5n+3-6×5n+19×5n-22×5n=5n+152-65n9-22=5n×5×25-65n9-4=5×195=19

(iv)
68n+1+1623n-21023n+1-78n=623n+1+1623n-21023n+1-723n=623n+3+1623n-21023n+1-723n
=6×23n23+1623n2-21023n21-723n=23n48+423n20-7=5213=4

Page No 2.12:

Question 8:

Solve the following equations for x:

(i) 72x+3=1

(ii) 2x+1=4x-3

(iii) 25x+3=8x+3

(iv) 42x=132

(v) 4x-1×0.53-2x=18x

(vi) 23x-7=256

Answer:

(i)
72x+3=172x+3=702x+3=02x=-3x=-32

(ii)
2x+1=4x-32x+1=22x-32x+1=22x-6x+1=2x-6x=7

(iii)
25x+3=8x+325x+3=23x+325x+3=23x+95x+3=3x+92x=6x=3

(iv)
42x=132222x=12524x×25=124x+5=204x+5=0x=-54

(v)
4x-1×0.53-2x=18x22x-1×123-2x=123x22x-1×2-13-2x=2-3x22x-2×22x-3=2-3x22x-2+2x-3=2-3x24x-5=2-3x4x-5= -3x7x= 5x= 57

(vi)
23x-7=25623x-7=283x-7=83x=15x=5



Page No 2.13:

Question 9:

Solve the following equations for x:

(i) 22x-2x+3+24=0

(ii) 32x+4+1=2.3x+2

Answer:

(i)
22x-2x+3+24=02x2-2x×23+222=02x2-2×2x×22+222=02x-222=02x-22=02x=22x=2

(ii)
32x+4+1=2.3x+23x+22-2.3x+2+1=03x+2-12=03x+2-1=03x+2=1=30x+2=0x=-2

Page No 2.13:

Question 10:

If 49392=a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.

Answer:

First find out the prime factorisation of 49392.

2493922246962123482617433087310297343749771

It can be observed that 49392 can be written as 24×32×73, where 2, 3 and 7 are positive primes.
49392=243273=a4b2c3a=2, b=3, c=7

Page No 2.13:

Question 11:

If 1176=2a3b7c, find a, b and c.

Answer:

First find out the prime factorisation of 1176.
21176258822943147749771

It can be observed that 1176 can be written as 23×31×72.
1176=233172=2a3b7c
Hence, a = 3, b = 1 and c = 2.

Page No 2.13:

Question 12:

Given 4725=3a5b7c, find
(i) the integral values of a, b and c
(ii) the value of 2-a3b7c

Answer:

(i) Given 4725=3a5b7c
First find out the prime factorisation of 4725.
347253157535255175535771
It can be observed that 4725 can be written as 33×52×71.
4725=3a5b7c=335271
Hence, a = 3, b = 2 and c = 1.

(ii)
When a = 3, b = 2 and c = 1,
2-a3b7c=2-3×32×71=18×9×7=638
Hence, the value of 2-a3b7c is 638.

Page No 2.13:

Question 13:

If a=xyp-1,b=xyq-1 and c=xyr-1, prove that aq-rbr-pcp-q=1.

Answer:

It is given that a=xyp-1,b=xyq-1 and c=xyr-1.
aq-rbr-pcp-q=xyp-1q-rxyq-1r-pxyr-1p-q=xq-ryp-1q-rxr-pyr-pq-1xp-qyp-qr-1=xq-rxr-pxp-qyp-1q-ryr-pq-1yp-qr-1=xq-r+r-p+p-qyp-1q-r+r-pq-1+p-qr-1=xq-r+r-p+p-qypq-q-pr+r+rq-r-pq+p+pr-p-qr+q=x0y0=1



Page No 2.24:

Question 1:

Assuming that x, y, z are positive real numbers, simplify each of the following:

(i) x-35

(ii) x3 y-2

(iii) (x-2/3y-1/2)2

(iv) (x)-2/3y4÷xy-1/2

(v) 243 x10y5z105

(vi) x-4y-105/4

(vii) 235672

Answer:

We have to simplify the following, assuming thatare positive real numbers

(i) Given

As x is positive real number then we have

Hence the simplified value of is

(ii) Given

As x and y are positive real numbers then we can write 

By using law of rational exponents we have 

Hence the simplified value of is

(iii) Given

As x and y are positive real numbers then we have 

By using law of rational exponents we have 

By using law of rational exponents we have


x-23y-122=1x23+23×1y12+12=1x43×1y22=1x43×1y=1x43y

Hence the simplified value of is .


 iv x-23 y4 ÷ xy-12=x12-23 y412 ÷ x × y-1212=x12×-23 × y4×12x12 × y-12×12    =x-13 × y2x12 × y-14by using the law of rational exponents, am ÷ an = am-n, we have     x-13-12 × y2+14=x-56 × y94            =y94x56v. 243  x10 y5 z105=243 × x10 × y5 × z1015=24315 × x1015 × y515 × z1015=3515 × x10×15 × y5×15 × z10×15=3 × x2 × y × z2=3x2yz2vi x-4y-1054=x-454y-1054=x-4×54y-10×54=x-5y-252=y252x5

(vii) 235672

235672=232+2+1672=232×232×231×672=23×23×231×672
=23×23×231×672=162493=5127203=512720312

 

Page No 2.24:

Question 2:

Simplify:

(i) (16-1/5)5/2

(ii) 32-35

(iii) (343)-23

(iv) (0.001)1/3

(v) (25)3/2×(243)3/5(16)5/4×(8)4/3

(vi) 258 ÷ 2513

(vii) 5-1×7252×7-47/2 × 5-2×7353×7-5-5/2

Answer:

(1) Given

By using law of rational exponents we have

Hence the value of is
(ii) 
32-35

=13235=13235=12535=125×35
=123=18

(iii) Given

Hence the value of is

(iv) Given

The value of is

(v) Given

Hence the value of is

(vi) Given. So,

By using the law of rational exponents

Hence the value of is

(vii) Given . So,

Hence the value of is

Page No 2.24:

Question 3:

Prove that:

(i) 3×5-3 ÷ 3-13 5×3×566=35

(ii) 93/2-3×50-181-1/2 = 15

(iii) 14-2-3×82/3×40+916-1/2 = 163

(iv) 21/2×31/3×41/410-1/5×53/5 ÷ 34/3×5-7/54-3/5×6=10

(v) 14 +(0.01)-1/2-(27)2/3 =32

(vi) 2n + 2n -12n+1-2n=32

(vii) 64125-2/3 + 12566251/4 + 25643 = 6516

(viii) 3-3×62×9852×1/253×(15)-4/3×31/3=282

(ix) (0.6)0-(0.1)-138-1323+-13-1=-32

Answer:

(i) We have to prove that

By using rational exponent we get,

Hence,

(ii) We have to prove that. So,

Hence,

(iii) We have to prove that

Now,

Hence,

(iv) We have to prove that. So,

Let

Hence,

(v) We have to prove that

Let

 

Hence,

(vi) We have to prove that . So,

Let

Hence,

(vii) We have to prove that. So let

By taking least common factor we get 

 

Hence,

(viii) We have to prove that. So,

Let

Hence,

(ix) We have to prove that. So,

Let

Hence,



Page No 2.25:

Question 4:

Show that:
(i) 11+xa-b+11+xb-a=1

(ii) xaa-bxaa+b÷xbb-axbb+aa+b=1

(iii) x1a-b1a-cx1b-c1b-ax1c-a1c-b=1

(iv) xa2+b2xaba+bxb2+c2xbcb+cxc2+a2xaca+c=x2a3+b3+c3

(v) xa-ba+bxb-cb+cxc-ac+a=1

(vi) xa-a-11a-1aa+1=x

(vii) ax+1ay+1x+yay+2az+2y+zaz+3ax+3z+x=1

(viii) 3a3ba+b3b3cb+c3c3ac+a=1

Answer:

(i)
11+xa-b+11+xb-a
=11+xaxb+11+xbxa=xbxb+xa+xaxa+xb=xb+xaxa+xb=1

(ii)
xaa-bxaa+b÷xbb-axbb+aa+b=xaa-bxaa+b×xbb+axbb-aa+b=xa2-abxa2+ab×xb2+abxb2-aba+b
=xa2-ab-a2-ab×xb2+ab-b2+aba+b=x-2ab×x2aba+b=x-2ab+2aba+b=x0a+b=1

(iii)
x1a-b1a-cx1b-c1b-ax1c-a1c-b=x1a-b×1a-cx1b-c×1b-ax1c-a×1c-b=x1a-b×1a-c+1b-c×1b-a+1c-a×1c-b=xb-c-a-c+a-ba-ba-cb-c=x0=1

(iv)
xa2+b2xaba+bxb2+c2xbcb+cxc2+a2xaca+c=xa2+b2-aba+bxb2+c2-bcb+cxc2+a2-aca+c=xa+ba2+b2-abxb+cb2+c2-bcxa+cc2+a2-ac=xa3+b3xb3+c3xa3+c3=x2a3+b3+c3

(v)
xa-ba+bxb-cb+cxc-ac+a=xa-ba+bxb-cb+cxc-ac+a=xa2-b2xb2-c2xc2-a2=xa2-b2+b2-c2+c2-a2=x0=1

(vi)
 xa-a-11a-1aa+1=xa-1a1a-1×aa+1=xa2-1aaa2-1=xa2-1a×aa2-1=x1=x

(vii)
ax+1ay+1x+yay+2az+2y+zaz+3ax+3z+x=ax+1-y-1x+yay+2-z-2y+zaz+3-x-3z+x=ax-yx+yay-zy+zaz-xz+x=ax2-y2+y2-z2+z2-x2=a0=1

(viii)
3a3ba+b3b3cb+c3c3ac+a=3a-ba+b3b-cb+c3c-ac+a=3a2-b2+b2-c2+c2-a2=30=1

Page No 2.25:

Question 5:

Question

Answer:

Let 2x=3y=12z=k
2=k1x,3=k1y,12=k1z
Now,
12=k1z22×3=k1zk1x2×k1y=k1zk2x+1y=k1z2x+1y=1z

Page No 2.25:

Question 6:

Question

Answer:

 Let 2x=3y=6-z=k
2=k1x, 3=k1y,6=k1-z
Now,
6=2×3=k1-zk1x×k1y=k1-zk1x+1y=k1-z1x+1y=1-z1x+1y+1z=0

Page No 2.25:

Question 7:

Question

Answer:

Let ax=by=cz=k
So, a=k1x,b=k1y,c=k1z
Thus, 
b2=ack1y2=k1xk1zk2y=k1x+1z2y=1x+1z
2y=z+xxzy=2zxz+x



Page No 2.26:

Question 8:

Question

Answer:

Let 3x=5y=75z=k
3=k1x, 5=k1y, 75=k1z52×3=k1zk1y2×k1x=k1zk2y×k1x=k1z
k2y+1x=k1z2y+1x=1z2x+yxy=1zz=xy2x+y

Page No 2.26:

Question 9:

If 27x =93x,  find x.

Answer:

We are given. We have to find the value of

Since

By using the law of exponents we get, 

On equating the exponents we get,

Hence,

Page No 2.26:

Question 10:

Find the values of x in each of the following:

(i) 25x÷2x =2205

(ii) (23)4=(22)x

(iii) 35x 532x=12527

(iv) 5x-2×32x-3 =135

(v) 2x-7×5x-4=1250

(vi) 432x+12=132

(vii) 52x+3=1

(viii) 13x=44-34-6

(ix) 35x+1=12527

Answer:

From the following we have to find the value of x

(i) Given

By using rational exponents

On equating the exponents we get,

The value of x is

(ii) Given

On equating the exponents 

Hence the value of x is

(iii) Given

Comparing exponents we have,

Hence the value of x is

(iv) Given

On equating the exponents of 5 and 3 we get,

And,

The value of x is

(v) Given

On equating the exponents we get 

And, 

Hence the value of x is

(vi) 
432x+12=132

22134x+12=12524x+13=2-5

On comparing we get, 
4x+13=-54x+1=-154x=-16x=-4

(vii) 52x+3=1

52x+3=502x+3=0x=-32

(viii) 13x=44-34-6

13x=224-34-613x=28-34-613x=256-81-613x=16913x=132

On comparing we get, 
x=2on squaring both sides we get, x=4

(ix) 35x+1=12527
35x+1=53335x+12=35-3

On comparing we get, 

x+12=-3x+1=-6x=-7

Page No 2.26:

Question 11:

Question

Answer:

 x=213+223
Cubing on both sides, we get
x3=213+2233x3=2133+2233+3×213×223213+223x3=2+22+3×21+23xx3=2+4+3×2xx3-6x=6

Page No 2.26:

Question 12:

Question

Answer:

9x+2=240+9x
9x×92=240+9x9x92-1=2409x81-1=2409x×80=2409x=332x=312x=1x=12
 8xx=8×1212=412=2

Page No 2.26:

Question 13:

Question

Answer:

 3x+1=9x-2
3x×3=9x923x×3=32x322=32x3434×3=32x3x35=32x-x
35=3x
Comparing both sides, we get
x = 5
So, 21+x=21+5=26=64

Page No 2.26:

Question 14:

If 34x=81-1 and 101y=0.0001, find the value of 2-x+4y.

Answer:

It is given that 34x=81-1 and 101y=0.0001.
Now,
34x=81-134x=34-13x4=3-14x=-1
And,
101y=0.0001101y=110000101y=1104101y=10-41y=-4y=-14
Therefore, the value of 2-x+4y is 21+4-14=20=1.

Page No 2.26:

Question 15:

If 53x=125 and 10y=0.001, find x and y.

Answer:

It is given that 53x=125 and 10y=0.001.
Now,
53x=12553x=533x=3x=1
And,
10y=0.00110y=1100010y=10-3y=-3
Hence, the values of x and y are 1 and −3, respectively.

Page No 2.26:

Question 16:

Solve the following equations:
(i) 3x+1=27×34
(ii) 42x=163-6y=82
(iii) 3x-1×52y-3=225
(iv) 8x+1=16y+2 and, 123+x=143y
(v) 4x-1×0.53-2x=18x

(vi) ab=ba1-2x, where a and b are distinct primes

Answer:

(i)
3x+1=27×343x+1=33×343x+1=33+43x+1=37x+1=7x=6

(ii)
42x=163-6y=8242x=82 and 163-6y=8242x=812×2 and 1613×-6y=812×2 42x=8 and 16-2y=824x=23 and 2-8y=234x=3 and -8y=3x=34 and y=-83

(iii)
3x-1×52y-3=2253x-1×52y-3=3×3×5×53x-1×52y-3=32×52x-1=2 and 2y-3=2x=3 and y=52

(iv)
8x+1=16y+2 and, 123+x=143y23x+1=24y+2 and 123+x=1223y23x+3=24y+8 and 123+x=126y3x+3=4y+8 and 3+x=6y
Now,
3+x=6yx=6y-3
Putting x = 6y − 3 in 3x-4y=5, we get
36y-3-4y=518y-9-4y=514y=14y=1
Putting y = 1 in x=6y-3, we get
x=6×1-3=3

(v)
4x-1×0.53-2x=18x22x-1×123-2x=123x22x-2×2-3-2x=2-3x22x-2-3+2x=2-3x4x-5=-3x7x=5x=57

(vi)
ab=ba1-2xab12=ab-1-2x12=-1-2x12=2x-132=2xx=34

Page No 2.26:

Question 17:

If a and b are distinct primes such that a6b-43=axb2y, find x and y.

Answer:

Given: a6b-43=axb2y
a6b-43=axb2ya6b-413=axb2ya6×13b-4×13=axb2ya2b-43=axb2yx=2  and y=-23

Page No 2.26:

Question 18:

If a and b are different positive primes such that

(i) a-1b2a2b-47÷a3b-5a-2b3=axby, find x and y.

(ii) a+b-1a-1+b-1=axby, find x + y + 2.

Answer:

(i)
a-1b2a2b-47÷a3b-5a-2b3=axbya-7b14a14b-28÷a3b-5a-2b3=axbya-7-14b14+28÷a3+2b-5-3=axbya-21b42÷a5b-8=axbya-21-5b42+8=axbya-26b50=axbyx=-26  and  y=50

(ii)
a+b-1a-1+b-1=axby1a+b1a+1b=axby1a+ba+bab=axby1ab=axbyab-1=axbya-1b-1=axbyx=-1  and  y=-1
Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.

Page No 2.26:

Question 19:

If 2x×3y×5z=2160, find x , y and z. Hence, compute the value of 3x×2-y×5-z.

Answer:

Given: 2x×3y×5z=2160
First, find out the prime factorisation of 2160.
2216021080254022703135345315551
It can be observed that 2160 can be written as 24×33×51.
Also,
2x×3y×5z=24×33×51x=4, y=3, z=1
Therefore, the value of 3x×2-y×5-z is 34×2-3×5-1=81×18×15=8140.

Page No 2.26:

Question 20:

If 1176=2a×3b×7c, find the values of a, b and c. Hence, compute the value of 2a×3b×7-c as a fraction.

Answer:

First find the prime factorisation of 1176.
21176258822943147749771
It can be observed that 1176 can be written as 23×31×72.
1176=2a3b7c=233172
So, a = 3, b = 1 and c = 2.
Therefore, the value of 2a×3b×7-c  is 23×31×7-2=8×3×149=2449



Page No 2.27:

Question 21:

Simplify:

(i) xa+bxca-bxb+cxab-cxc+axbc-a

(ii) lmxlxm×mnxmxn×nlxnxl

Answer:

(i)
xa+bxca-bxb+cxab-cxc+axbc-a=xa+ba-bxca-bxb+cb-cxab-cxc+ac-axbc-a=xa+ba-bxca-bcxb+cb-cxab-acxc+ac-axbc-ab=xa2-b2xb2-c2xc2-a2xca-bc+ab-ac+bc-ab=xa2-b2+b2-c2+c2-a2xca-bc+ab-ac+bc-ab=x0x0=1

(ii)
xlxmlm×xmxnmn×xnxlnl=xlxm1ml×xmxn1mn×xnxl1nl=xl-m1ml×xm-n1mn×xn-l1nl=xl-mml×xm-nmn×xn-lnl=xl-mml+m-nmn+n-lnl=xln-mn+lm-nl+nm-lmnml=x0=1

Page No 2.27:

Question 22:

Show that: a+1bm×a-1bnb+1am×b-1an=abm+n

Answer:


a+1bm×a-1bnb+1am×b-1an=ab+1bm×ab-1bnab+1am×ab-1an=ab+1bab+1am×ab-1bab-1an=ab+1b×aab+1m×ab-1b×aab-1n=abm×abn=abm×abn=abm+n

Page No 2.27:

Question 23:

(i) If a=xm+nyl, b=xn+lym and c=xl+myn, prove that am-nbn-lcl-m=1.

(ii) If x=am+n, y=an+l and z=al+m, prove that xmynzl=xnylzm.

Answer:

(i) Given: a=xm+nyl, b=xn+lym and c=xl+myn
Putting the values of a, b and c in am-nbn-lcl-m, we get
am-nbn-lcl-m=xm+nylm-nxn+lymn-lxl+mynl-m=xm+nm-nylm-nxn+ln-lymn-lxl+ml-mynl-m=xm2-n2xn2-l2xl2-m2ylm-lnymn-mlynl-nm=xm2-n2+n2-l2+l2-m2ylm-ln+mn-ml+nl-nm=x0y0=1

(ii) Given: x=am+n, y=an+l and z=al+m
Putting the values of x, y and z in xmynzl, we get
xmynzl=am+nman+lnal+ml=am2+nman2+lnal2+lm=am2+n2+l2+nm+ln+lm
Putting the values of x, y and z in xnylzm, we get
xnylzm=am+nnan+l lal+mm=amn+n2anl+l2 alm+m2=amn+n2+nl+l2+lm+m2
So, xmynzl = xnylzm



Page No 2.28:

Question 1:

Write 625-1/4 in decimal form.

Answer:

We have to writein decimal form. So,

625-14=162514=15414

Hence the decimal form of is

Page No 2.28:

Question 2:

State the product law of exponents.

Answer:

State the product law of exponents.

If is any real number and , are positive integers, then

By definition, we have 

(factor) ( factor)

to factors

Thus the exponent "product rule" tells us that, when multiplying two powers that have the same base, we can add the exponents. 

Page No 2.28:

Question 3:

State the quotient law of exponents.

Answer:

The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. If a is a non-zero real number and m, n are positive integers, then

We shall divide the proof into three parts 

(i) when

(ii) when

(iii) when

Case 1 

When

We have 

aman=a×a×a....to m factorsa×a×a....to n factorsaman=a×a×a....to (m-n) factorsaman=am-n

Case 2 

When

We get

Cancelling common factors in numerator and denominator we get,

By definition we can write 1 as

Case 3 

When

In this case, we have 

Hence, whether, or,

Page No 2.28:

Question 4:

State the power law of exponents.

Answer:

The "power rule" tell us that to raise a power to a power, just multiply the exponents. 

If a is any real number and m, n are positive integers, then

We have,

factors

factors

Hence,



Page No 2.29:

Question 5:

For any positive real number x, find the value of

xaxba+b×xbxcb+c×xcxac+a

Answer:

We have to find the value of L =

By using rational exponents, we get

By using rational exponents we get 

By definition we can write as 1

Hence the value of expression is .

Page No 2.29:

Question 6:

Write the value of 5(81/3+271/3)31/4.

Answer:

We have to find the value of. So,

By using rational exponents we get 

Hence the simplified value of is

Page No 2.29:

Question 7:

Simplify 625-1/2-1/42

Answer:

We have to simplify. So,

Hence, the value of is

Page No 2.29:

Question 8:

For any positive real number x, write the value of

xab1abxbc1bcxca1ca

Answer:

We have to simplify. So,

By using rational exponents, we get

Hence the value of is

Page No 2.29:

Question 9:

If (− 1)3 = 8, What is the value of (+ 1)2 ?

Answer:

We have to find the value of , where 

Consider

By equating the base, we get

By substituting in

Hence the value of is .

Page No 2.29:

Question 10:

If 24 × 42 =16x, then find the value of x.

Answer:

We have to find the value of x provided

So,

By equating the exponents we get

Hence the value of x is .

Page No 2.29:

Question 11:

If 3x-1 = 9 and 4y+2 = 64, what is the value of xy ?

Answer:

We have to find the value of for

So,

By equating the exponent we get

Let’s take

By equating the exponent we get

By substituting in we get

Hence the value of is

Page No 2.29:

Question 12:

Write the value of 73×493.

Answer:

We have to find the value of . So,

By using law rational exponents we get,

Hence the value of is

Page No 2.29:

Question 13:

Write 19-1/2×(64)-1/3 as a rational number.

Answer:

We have to find the value of . So,

Hence the value of the value of is .

Page No 2.29:

Question 14:

Write the value of 125×273.

Answer:

We have to find the value of 125×273. So,


125×273=53×333=5×3=15

Hence the value of the value of is .

Page No 2.29:

Question 1:

The value of 2-3 (2-3)33 is

(a) 5

(b) 125

(c) 1/5

(d) -125

Answer:

We have to find the value of. So,

The value of is 125

Hence the correct choice is

Page No 2.29:

Question 2:

(256)0.16 × (256)0.09

(a) 4
(b) 16
(c) 64
(d) 256.25

Answer:

We have to find the value of. So,

By using law of rational exponents

we get

The value of is 4

Hence the correct choice is .

Page No 2.29:

Question 3:

If 102y = 25, then 10-y equals

(a) -15

(b) 150

(c) 1625

(d) 15

Answer:

We have to find the value of

Given that, therefore,

Hence the correct option is .

Page No 2.29:

Question 4:

The value of x − yx-y when x = 2 and y = −2 is

(a) 18
(b) −18
(c) 14
(d) −14

Answer:

Given

Here

By substituting in we get 

The value of is – 14

Hence the correct choice is .



Page No 2.30:

Question 5:

The product of the square root of x with the cube root of x is
(a) cube root of the square root of x
(b) sixth root of the fifth power of x
(c) fifth root of the sixth power of x
(d) sixth root of x

Answer:

We have to find the product (say L) of the square root of x with the cube root of x is. So, 

=x3+26=x56

The product of the square root of x with the cube root of x is

Hence the correct alternative is

Page No 2.30:

Question 6:

If 9x+2 = 240 + 9x, then x =

(a) 0.5
(b) 0.2
(c) 0.4
(d) 0.1

Answer:

We have to find the value of

Given

By equating the exponents we get 

Hence the correct alternative is .

Page No 2.30:

Question 7:

The seventh root of x divided by the eighth root of x is
(a) x

(b) x

(c) x56

(d) 1x56

Answer:

We have to find he seventh root of x divided by the eighth root of x, so let it be L. So, 

The seventh root of x divided by the eighth root of x is

Hence the correct choice is .

Page No 2.30:

Question 8:

The square root of 64 divided by the cube root of 64 is

(a) 64
(b) 2
(c) 12
(d) 642/3

Answer:

We have to find the value of

So,

The value of is

Hence the correct choice is .

Page No 2.30:

Question 9:

Which of the following is (are) not equal to 561/5-1/6 ?
(a) 5615-36 

(b) 1561/51/6

(c) 651/30

(d) 56-1/30

Answer:

We have to find the value of

So,

Hence the correct choice is .

Page No 2.30:

Question 10:

When simplified (x-1+y-1)-1 is equal to

(a) xy

(b) x+y

(c) xyx+y

(d) x+yxy

Answer:

We have to simplify

So,

The value of is

Hence the correct choice is .

Page No 2.30:

Question 11:

If 8x+1 = 64 , what is the value of 32x+1 ?

(a) 1
(b) 3
(c) 9
(d) 27

Answer:

We have to find the value of provided

So,

Equating the exponents we get

By substitute in we get 

The real value of is

Hence the correct choice is .

Page No 2.30:

Question 12:

If o <y <x, which statement must be true?

(a) x-y=x-y

(b) x + x = 2x

(c) xy=yx

(d) xy =xy

Answer:

We have to find which statement must be true?

Given

Option (a) :

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong. 

Option (b) : 

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong.

Option (c) : 

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong. 

Option (d) : 

Left hand side: 

Right Hand side:

Left hand side is equal to right hand side 

The statement is true.

Hence the correct choice is .

Page No 2.30:

Question 13:

If x is a positive real number and x2 = 2, then x3 =

(a) 2

(b) 22

(c) 32

(d) 4

Answer:

We have to find provided. So,

By raising both sides to the power

By substituting in we get

The value of is

Hence the correct choice is .

Page No 2.30:

Question 14:

If (23)2 = 4x, then 3x =

(a) 3
(b) 6
(c) 9
(d) 27

Answer:

We have to find the value ofprovided

So,

By equating the exponents we get

By substituting in we get 

The value of is

Hence the correct choice is



Page No 2.31:

Question 15:

If 10x = 64, what is the value of 10x2+1 ?

(a) 18
(b) 42
(c) 80
(d) 81

Answer:

We have to find the value of provided

So,

By substituting we get 

Hence the correct choice is .

Page No 2.31:

Question 16:

If xx1.5=8x-1 and x > 0, then x =

(a) 24

(b) 22

(c) 4

(d) 64  
 

Answer:

For, we have to find the value of x.

So,

By raising both sides to the power we get

The value of is

Hence the correct alternative is

Page No 2.31:

Question 17:

If g = t2/3+4t-1/2, What is the value of g when t = 64?

(a) 212

(b) 332

(c) 16

(d) 25716

Answer:

Given.We have to find the value of

So,

The value of is

Hence the correct choice is

Page No 2.31:

Question 18:

If x-2 = 64, then x1/3+x0 =

(a) 2
(b) 3
(c) 3/2
(d) 2/3

Answer:

We have to find the value ofif

Consider,

Multiply on both sides of powers we get 

By taking reciprocal on both sides we get,

Substituting in we get

By taking least common multiply we get 

Hence the correct choice is .

Page No 2.31:

Question 19:

If 4x - 4x-1 = 24, then (2x)x equals

(a) 55

(b) 5

(c) 255

(d) 125
 

Answer:

We have to find the value of if

So,

Taking as common factor we get 

By equating powers of exponents we get 

By substituting in we get

Hence the correct choice is

Page No 2.31:

Question 20:

When simplified -127-2/3 is

(a) 9

(b) −9

(c) 19

(d) -19

Answer:

We have to find the value of

So,

Hence the correct choice is .

Page No 2.31:

Question 21:

Which one of the following is not equal to 83-1/2 ?

(a) 23-1/2

(b) 8-1/6

(c) 1(83)1/2

(d) 12

Answer:

We have to find the value of

So, 

Also,

Hence the correct alternative is .

Page No 2.31:

Question 22:

Which one of the following is not equal to 1009-3/2 ?

(a) 91003/2

(b) 110093/2

(c) 310×310×310

(d) 1009×1009×1009

Answer:

We have to find the value of

So,

Since, is equal to ,,.

Hence the correct choice is

Page No 2.31:

Question 23:

When simplified (256) -(4-3/2) is  

(a) 8

(b) 18

(c) 2

(d) 12

Answer:

Simplify


256-4-32=256-2-3
 

Hence the correct choice is .



Page No 2.32:

Question 24:

5n+2-6×5n+113×5n-2×5n+1 is equal to

(a) 53

(b) -53

(c) 35

(d) -35


 

Answer:

We have to simplify

Taking as a common factor we get

Hence the correct alternative is

Page No 2.32:

Question 25:

If a, b, c are positive real numbers, then a-1b×b-1c×c-1a is equal to

(a) 1

(b) abc

(c) abc

(d) 1abc
 

Answer:

We have to find the value of when a, b, c are positive real numbers.

So,

Taking square root as common we get 

a-1b×b-1c×c-1a=ba×cb×aca-1b×b-1c×c-1a=1

Hence the correct alternative is .

Page No 2.32:

Question 26:

If32x-8225=535x, then x =

(a) 2
(b) 3
(c) 5
(d) 4

Answer:

We have to find the value of provided

So,

By cross multiplication we get 

By equating exponents we get 

And 

Hence the correct choice is

Page No 2.32:

Question 27:

, then x =

(a) 2
(b) 3
(c) 4
(d) 1

Answer:

We have to find value of provided

So,

Equating exponents of power we get

Hence the correct alternative is

Page No 2.32:

Question 28:

The value of 8-4/3÷2-21/2 is

(a) 12

(b) 2

(c) 14

(d) 4

Answer:

Find the value of

Hence the correct choice is .

Page No 2.32:

Question 29:

If a, b, c are positive real numbers, then 3125a10b5c105 is equal to

(a) 5a2bc2

(b) 25ab2c

(c) 5a3bc3

(d) 125a2bc2

Answer:

Find value of.

3125a10b5c105=5a2bc2

Hence the correct choice is .

Page No 2.32:

Question 30:

The value of 64-1/3 (641/3-642/3), is

(a) 1

(b) 13

(c) −3

(d) −2

Answer:

Find the value of

So,

Hence the correct statement is.

Page No 2.32:

Question 31:

If 5n=125, then =

(a) 25

(b) 1125

(c) 625

(d) 15

Answer:

We have to find provided

So,

Substitute in to get

Hence the value of is

The correct choice is

Page No 2.32:

Question 32:

If (16)2x+3 =(64)x+3, then 42x-2 =

(a) 64

(b) 256

(c) 32

(d) 512

Answer:

We have to find the value ofprovided

So,

Equating the power of exponents we get

The value of is 

Hence the correct alternative is

Page No 2.32:

Question 33:

If a, m, n are positive ingegers, then anmmnis equal to

(a) amn

(b) a

(c) am/n

(d) 1

Answer:

Find the value of .

So,

Hence the correct choice is

Page No 2.32:

Question 34:

If 2-m×12m=14, then 114(4m)1/2+15m-1 is equal to

(a) 12

(b) 2  

(c) 4

(d) -14

Answer:

We have to find the value ofprovided

Consider,

Equating the power of exponents we get 

By substituting we get 

Hence the correct choice is .



Page No 2.33:

Question 35:

If x = 2 and y = 4, then xyx-y+yxy-x =

(a) 4

(b) 8

(c) 12

(d) 2

Answer:

We have to find the value of if,

Substitute,into get,

Hence the correct choice is .

Page No 2.33:

Question 36:

The value of m for which 172-2-1/31/4=7m, is

(a) -13

(b) 14

(c) −3

(d) 2

Answer:

We have to find the value of for

By using rational exponents

7-13=7m

Equating power of exponents we get

Hence the correct choice is .

Page No 2.33:

Question 37:

If 2m+n2n-m=163p3n=81 and a=21/10, thena2m+n-p(am-2n+2p)-1=

(a) 2

(b) 14

(c) 9

(d) 18

Answer:

Given :  3p3n=81 and  
To find :  

Find : 
By using rational components We get

By equating rational exponents we get 

Now,   =a2m+n-p.am-2n+2p  we get
=a2m+n-p+m-2n+2p=a3m-n+pNow putting value of a = 2110 we get, =23m-n+p10=26-n+p10

Also, 3p3n=81
3p-n=34
On comparing LHS and RHS we get, p - n = 4.
Now, 
= a3m - n + p
=26+(p-n)10=26+410=21010=21=2


So, option (a) is the correct answer.

 

Page No 2.33:

Question 38:

The value of 23+222/3+(140-19)1/22,  is

(a) 196

(b) 289

(c) 324

(d) 400

Answer:

We have to find the value of

By using the identity we get,

Hence correct choice is .

Page No 2.33:

Question 39:

If 2n=1024,  then 32n4-4=

(a) 3

(b) 9

(c) 27

(d) 81

Answer:

We have to find

Given

Equating powers of rational exponents we get 

Substituting in we get 

Hence the correct choice is .

Page No 2.33:

Question 40:

If 35x×812×656132x=37, then x =

(a) 3

(b) −3

(c) 13

(d) -13

Answer:

We have to find the value of x provided

So,

By using law of rational exponents we get

By equating exponents we get

Hence the correct choice is .



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