LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential...
Transcript of LOGARITHMS AND INDICES - TestlabzMath Class IX 2 Question Bank Q.1. (C) By converting to exponential...
Math Class IX 1 Question Bank
LOGARITHMS AND
INDICES Q.1. (A) Convert each of the following to logarithmic form.
(i) 2552
= (ii) 27
13 3
=− (iii) 4(64)3
1
= (iv) 160
=
(v) 0.01102
=− (vi)
4
14
1=
−
Ans. We know that logb
aa x b x= ⇒ =
(i) 2552= ∴ 225log5 = (ii)
27
13 3
=− ∴ 3
27
1log3 −=
(iii) 4)64( 3
1
= ∴ 3
14log64 = (iv) 16
0= ∴ 01log6 =
(v) 01.0102
=− ∴ 2)01.0(log10 −= (vi)
4
14 1
=−
∴ 14
1log4 −=
Q.1. (B) Convert each of the following to exponential form.
(i) 481log3 = (ii) 3
24log8 = (iii) 3
8
1log2 −= (iv) 2(0.01)log10 −=
(v) 15
1log5 −=
(vi) loga 1 = 0
Ans.
(i) 481log3 = 43 81∴ = (ii) 3
24log8 =
2
3 48∴ =
(iii) 38
1log2 −= 3 1
28
−∴ = (iv) 2)01.0(log10 −= 210 0.01−
∴ =
(v) 15
1log5 −=
1 1
55
−∴ = (vi) 01log =a 0 1a∴ =
Math Class IX 2 Question Bank
Q.1. (C) By converting to exponential form find the value of each of the
following.
(i) 64log2 (ii) 32log8 (iii) 9
1log3 (iv) (16)log0.5
(v) (0.125)log2 (vi) 7log7
Ans. (i) Suppose x=64log2 , then 222222642 ×××××==x ⇒
622 =
x ∴
6=x Hence, 664log2 =
(ii) Suppose ,32log8 x= then 328 =x ⇒ 53
22 =x ∴ 53 =x ⇒
3
5=x
Hence, 3
532log8 =
(iii) Suppose ,9
1log3 x= then 2
23
3
1
9
13
−===
x ∴ 2−=x . Hence,
29
1log3 −=
(iv) Suppose ,)16(log 5.0 x= then 16)5.0( =x ⇒ 2222
2
1×××=
x
⇒ 4
22 =−x
∴ 44 −=⇒=− xx . Hence, 4)16(log 5.0 −=
(v) Suppose ,)125.0(log2 x= then 3
2
1
8
1
1000
125.0125.02 ====
x
⇒ 3
22−
=x
3−=∴ x . Hence, 3)125.0(log2 −=
(vi) Suppose ,7log7 x= then 1777 ==
x 1=∴ x
Hence, 17log7 =
Q.2. Find the value of x, when :
(i) x = −2log 2 (ii) x =log 9 1 (iii) x=9log 243
(iv) x =3log 0 (v) x= −4log 32 4 (vi) x − =3
log ( 1) 2
(vii) x − =2
7log (2 1) 2 (viii) x =
3log 64
2
Math Class IX 3 Question Bank
Ans. (i) 2log2 −=x ∴ x=−2
2 ⇒ 4
1
22
1
2
12
=×
==x . Hence, .4
1=x
(ii) 19log =x ∴ 91
=x 9=⇒ x . Hence, 9=x
(iii) x=243log9 ∴ 2439 =x ⇒ 2(3 ) 3 3 3 3 3x
= × × × × ⇒ 5233 =
x
⇒ 52 =x ⇒ 5.22
5==x . Hence, 2.5x =
(iv) 0log3 =x
⇒ 130
=⇒= xx Hence, 1=x )1( 0=x∵
(v) 432log4 −= x
∴ 324 4=
−x ⇒ 542)22( =×
−x ⇒ 5422)2( =
−x ⇒ 58222 =
−x
On comparing both sides 582 =−x
⇒ 2
13132 =⇒= xx . Hence,
13
2x =
(vi) 2)1(log3
=−x
∴ 1)3( 2−= x ⇒
122(3 ) 1 3 1x x= − ⇒ = − ∴ 413 =+=x . Hence, 4x =
(vii) 2)12(log2
7 =−x
∴ 12722
−= x ⇒ 12492
−= x ⇒ 14922
+=x ⇒ 5022
=x
⇒ 252
502==x ⇒ 25 5x = = ±
Hence, 5x = ±
(viii) 2
364log =x
∴ 642
3
=x ⇒ 3/233/2)4()64( ==x 1644
23
23
===
×
Hence, .16=x
Math Class IX 4 Question Bank
Q.3. If x p=10log and y q=10log , show that p qxy
+= (10) .
Ans. ∵ px =10log
∴ xp
=10 ...(i)
Similarly, qy =10log
yq
=10 ...(ii)
Multiplying equation (i) by (ii), we get
qpqpxy
+=×= )10(1010 Hence, qp
xy+
= )10(
Q.4. Given x a y b= =10 10log , log ,
(i) Write down a+110 in terms of x. (ii) Write down b2
10 in terms of y.
(iii) If P a b= −10log 2 , Express P in terms of x and y.
Ans. 10 10log , logx a y b= =
∴ xb
=10 and yb
=10
(i) 11101010 ×=
+ aa xx 1010 =×= )10Put ( xa
=
(ii) bbb101010
2×= )10( y
b=
2yyy =×=
(iii) baP −= 2log10 ∴ Pba
=−2
10
⇒ 210 10
a bP = ÷
ba 10)10( 2÷=
yx ÷=2 (Put x
a=10 and )10 y
b=
2x
Py
=
Q.5. Evaluate the following without using log tables :
(i) log42
1log82log5 −+ (ii) log182log3log25log8 −++
(iii) 243
32log
9
52log
16
75log +− (iv) −
3 15log2 + log25 + log49 log28
2 2
(v) 30
1loglog36
2
1log52log2 −−+ (vi) −log (1.2) + 2log(0.75) log(6.75)
Math Class IX 5 Question Bank
Ans. (i) 4log2
18log5log2 −+ 2
1
2)4log(8log)5log( −+= ( log log )
nm n m=∵
2
825log2log8log25log
×=−+=
[∵ mnnm logloglog =+ and log log log ]m
m nn
− =
2log100 log10 2log10 2= = = =
(ii) 18log3log225log8log −++2log8 log 25 log(3) log18= + + −
18log9log25log8log −++=
××=
18
9258log
2log100 log10 2log10 2= = = =
(iii) 243
32log
9
5log2
16
75log +−
243
32log
9
5log
16
75log
2
+
−= (∵ )loglog
naa mmn =
243
32log
9
5
9
5log
16
75log +×−=
243
32log
81
25log
16
75log +−=
75
3216log log25 243
81
= + (∵ )logloglog nmn
maaa −=
243
32log
25
81
16
75log +×=
243
32log
25
81
16
253log +×
×=
243
32log
1
81
16
13log +×
×=
243
32log
16
243log +=
×=
243
32
16
243log 2log)21log( =×=
Math Class IX 6 Question Bank
(iv) 28log49log2
125log
2
32log5 −++
28log)49log()25log(2log 2
1
2
3
5−++= 28log7log)5log(32log
3−++=
28log7log125log32log −++=32 125 7
log log100028
× ×= =
3log10 3log10 3= = =
(v) 30
1log36log
2
15log2log2 −−+
30
1log)36log(5log2log 2
1
2−−+= ]30log1[log6log5log4log −−−+=
30log1log6log5log4log +−−+=4 5 30
log log1006 1
× ×= =
×
2log10 2log10 2= = =
(vi) )75.6log()75.0log(2)2.1log( −+
)75.6log()75.0log()2.1log( 2−+= log(1.2) log(0.5625) log(6.75)= + −
1.2 0.5625
log6.75
×=
12 5625 100log
10 10000 675
× ×=
× ×11.0log
10
1log −===
Q.6. Express each of the following as a single logarithm :
(i) 23log(1.5)log36log82log 10101010 −−+ (ii) 3log22log51 +−
(iii) 12log32log52log 1010102 +−+ (iv) 52log9log2
12 1010 −+
(v) 12log62log8log4
19log
2
110101010 −++
(vi)
−
+
91
55log
77
130log
13
112log 101010
Ans. (i) 2log3)5.1(log36log8log2 10101010 −−+
3
1010102
10 2log)5.1(log36log)8(log −−+=
8log5.1log36log64log 10101010 −−+=
×
××=
×
×=
815
103664log
85.1
3664log 1010 192log10=
Math Class IX 7 Question Bank
(ii) 2log35log21 +−32
2log5log10log +−= [∵ ]110log =a
8log25log10log +−=
8log)25log10(log +−=
8log25
10log −= 8
25
10log ×=
16log
5=
(iii) 12log3log25log2 101010 +−+
10log2log)3(log)5(log 10102
102
10 +−+= ]110log[ 10 =∵
10log2log9log25log 10101010 +−+= 1125log2
10925log 1010 =
××=
(iv) 5log29log2
12 1010 −+
210
2
1
1010 )5(log)9(log100log −+= [∵ ]2100log10 =
25log3log100log 101010 −+=
12log25
3100log 1010 =
×=
(v) 12log6log281log4
19log
2
110101010 −++
12log)6(log)81(log)9(log 102
104
1
102
1
10 −++=
12log36log3log3log 10101010 −++=
27log12
3633log 1010 =
××=
(vi)
−
+
91
55log
77
130log
13
11log2 101010
−
+
=
91
55log
77
130log
13
11log 1010
2
10
−
+=
91
55log
77
130log
169
121log 101010
−
×=
91
55log
77
130
169
121log 1010
Math Class IX 8 Question Bank
91
5577
130
169
121
log10
×
=55
91
77
130
169
121log10 ××=
2log10=
Q.7. Evaluate the following without using log tables :
(i) log27
log81 (ii)
log32
log128
(iii) 3log
log27 (iv)
log27
log3log9 −
Ans. (i) 3
4
3log3
3log4
3log
3log
27log
81log3
4
===
(ii) 5
7
2log5
2log7
2log
2log
32log
128log5
7
===
(iii) 61
23
2
1
3
3log2
1
3log3
3log
3log
3log
27log
2
1
3
=×
====
(iv) 3
1
3log3
3log
3log3
3log3log2
3log
3log3log
27log
3log9log3
2
==−
=−
=−
Q.8. Solve for x :
(i) x − =10log ( 10) 1 (ii) x − =2
log ( 21) 2
(iii) x x− + + =log ( 2) log ( 2) log5
(iv) x x+ + − = +log ( 5) log ( 5) 4log2 2log3
(v) x x+ − − =log ( 4) log ( 4) log2
(vi) x x+ − − =log ( 3) log ( 3) 1
(vii) x=log81
log27 (viii) x=
log128
log32
Ans. (i) 1)10(log10 =−x
⇒ 10log)10(log 1010 =−x )110log( =∵
⇒ 2010101010 =⇒+=⇒=− xxx
(ii) 2)21(log 2=−x
Math Class IX 9 Question Bank
⇒ 100log)21(log 102
10 =−x )2log( 100 =∵
⇒ 100212
=−x
⇒ 1212110022
=⇒+= xx ⇒ 11=x
(iii) 5log)2(log)2(log =++− xx
⇒ 5log)2()2(log =+×− xx [∵ ]logloglog mnnm aaa =+
⇒ 2log ( 4) log5x − =
⇒ 94554222
=⇒+=⇒=− xxx ⇒ 3)3()( 22=⇒= xx
(iv) 3log22log4)5(log)5(log +=−++ xx
⇒ 3log22log4)5()5(log +=−×+ xx [∵ ]logloglog mnnm aaa =+
⇒ 24223log2log])5()[(log +=−x
⇒ 2log ( 25) log 16 log9x − = +
⇒ )916(log)25(log2
×=−x [∵ ]logloglog mnnm aaa =+
⇒ 144log)25(log 2=−x ⇒ 14425
2=−x ⇒ 25144
2+=x
⇒ 13)13()(169222
=⇒=⇒= xxx
(v) 2log)4(log)4(log =−−+ xx ⇒ 2log)4(
)4(log =
−
+
x
x
⇒ )4(2424
4−=+⇒=
−
+xx
x
x
⇒ 482824 −−=−⇒−=+ xxxx ⇒ 1212 =⇒−=− xx
(vi) 1)3(log)3(log =−−+ xx
⇒ 1)3(log)3(log 1010 =−−+ xx
⇒ 1)3(
)3(log10 =
−
+
x
x [∵ ]logloglog
n
mnm aa =−
⇒ 10)3(
)3(10log
)3(
)3(log 1010 =
−
+⇒=
−
+
x
x
x
x
⇒ )3(10)3( −=+ xx ⇒ 3 10 ( 3)x x+ = −
⇒ 3 10 30 10 30 3x x x x+ = − ⇒ − = − −
⇒ 9
33339339 =⇒=⇒−=− xxx
⇒ 11 2
33 3
x x= ⇒ =
Math Class IX 10 Question Bank
(vii) x=27log
81log⇒
)333(log
)3333(log
27log
81log
××
×××=⇒= xx
⇒ 3
4
3log3
3log4
3log
3log3
4
==⇒= xx
(viii) x=32log
128log
⇒ )22222(log
)2222222(log
32log
128log
××××
××××××=⇒= xx ⇒
2log5
2log7
2log
2log5
7
=⇒= xx
⇒ 5
21
5
7=⇒= xx
Q.9. Given log x m n= + and log y m n= − , express the value of x
y2
10log in
terms of m and n.
Ans. nmx +=log (Given) ...(i)
nmy −=log (Given) ...(ii)
2
2
10log log10 log
xx y
y= − (∵ )logloglog nm
n
maaa −=
2loglog10log yx −+=
yx log2log10log −+=
yx log2log1 −+= ]110log[ =∵
Putting the value of nmx +=log and nmy −=log , we get
)(2)(1 nmnm −−++= nmnmnm 31221 +−=+−++=
Math Class IX 11 Question Bank
Q.10. If a=10log 2 and b=10log 3 ; express each of the following in terms of ‘a’
and ‘b’:
(i) log12 (ii) log2.25 (iii) log5.4 (iv) log60
(v) 8
1log3 (vi) 18log (vii)
4
9log
Ans. a=2log10 ...(i)
b=3log10 ...(ii)
(i) 12log )32(log)322(log 2×=××=
3log2log2
+= 3log2log2 += ]loglog[ mnm an
a =∵
3log2log2 1010 +=
ba +×= 2 [Putting the value of a=2log10 and ]3log10 b=
ba += 2
(ii) 25.2log425
925log
100
225log
×
×==
2
2
3log
4
9log
==
]2log3[log22
3log2 −== )(2]2log3[log2 1010 ab −=−=
[Putting ]2log,3log 1010 ab ==
ab 22 −= 2( )b a⇒ −
(iii) 10log54log10
54log4.5log −== 10log)3332(log 10−×××=
10log)32(log 103
10 −×= 13log2log3
1010 −+= ]110[log10 =
13log32log 1010 −+= 1313 −+=−×+= baba
(iv) )3210(log60log 10 ××=
3log2log10log 101010 ++= ]logloglog[log 10101010 nmllmn ++=
ba ++= 1 [Putting a=2log10 and ]3log10 b=
1++= ba
Math Class IX 12 Question Bank
(v) 4
4
8
25log
8
25log
8
13log 10 ×==
22222
100log
32
100log 1010
××××==
51010510 2log100log
2
100log −== [∵ ]logloglog nm
n
maaa −=
2log52 10−= a×−= 52 [Putting ]2log10 a=
a52 −=
(vi) 2
1
)18(log18log = 2
1
)233(log ××=
)23(log2
1)23(log
22
1
2×=×= ]2log3[log
2
1 2+=
]2log3log2[2
1+= 2log
2
13log2
2
1+×=
2log2
13log2
2
1+×= 2log
2
13log +=
+=
2
ab
(vii)
2
2
3log
22
33log
4
9log
=
×
×=
32log 2 [log3 log 2] 2( )
2b a= = − = −
Q.11. If 0.3010,log2 = find the value of .243
32log
9
52log
16
75log
+−
Ans. 243
32log
9
5log2
16
75log +−
243
32log
9
5log
16
75log
2
+
−=
75 25 32
log log log16 81 243
= − +81
25log
243
32log
16
75log −+=
75 32 25
log log16 243 81
= × −
××=
25
81
243
32
16
75log
3010.02log ==
Q.12. If 0.9030,log8 = find the value of :
(i) log4 (ii) 32log (iii) (0.125)log
Ans. 9030.08log =∵ ⇒ 9030.0)2(log 3= ⇒ 9030.02log3 =
∴ 3010.03
9030.02log ==
(i) 2log22log4log 2== 6020.0)3010.0(2 ==
Math Class IX 13 Question Bank
(ii) 52
1
2log2
1)32(log32log ==
)3010.0(2
52log
2
5== 7525.01505.05 =×=
(iii)
==
8
1log
1000
125log)125.0(log
9030.0)3010.0(32log3)2(log2
1log
33
−=−=−==
=
−
Q.13. If 1.4313log27 = , find the value of :
(i) log9 (ii) log30
Ans. 4313.127log = ⇒ 4313.13log 3= ⇒ 4313.13log3 =
⇒ 477.03
4313.13log ==
(i) 9542.0)4771.0(23log23log9log 2====
(ii) 4771.114771.010log3log)103(log30log =+=+=×=
Q.14. Show that log3log2log13)2(1log ++=++
Ans. 3log2log1log)321(log6log)321(log ++=××==++ Hence, proved.
Q.15. (i) If log logm n m n+ = +log ( ) , show that n
mn
=− 1
(ii) If log loga b
a b+
= +
1log ( ),
2 2 show that a b ab+ =
1( ) .
2
Ans. (i) nmnm loglog)(log +=+
⇒ )(log)(log nmnm ×=+
∴ mnnm =+
⇒ nmnm −=−
⇒ (1 )m n n− = −
⇒ 1)1(1 −
=−−
−=
−
−=
n
n
n
n
n
nm . Hence,
1−=
n
nm
Math Class IX 14 Question Bank
(ii) )log(log2
1
2log ba
ba+=
+⇒ 2
1
)(log)(log2
1
2log abab
ba==
+
⇒ log log2
a bab
+ =
Comparing both sides, we get
abba
=+
2 or abba =+ )(
2
1. Hence, .)(
2
1abba =+
Q.16. If log loga b a b+ = +log ( ) , find a in terms of b.
Ans. baba loglog)(log +=+
⇒ abba log)(log =+ [∵ ]logloglog nmmn aaa +=
⇒ abba =+ ⇒ baba −=−
⇒ baab −=+− ⇒ bba −=−− )1( ⇒1
)1(−
=⇒=−b
babba
Q.17. If a b
mbc ca
= =
2 2
l log , log and c
nab
=
2
log,, find the value of l m n+ + .
Ans. ab
cn
ca
bm
bc
al
222
log,log,log ===
ab
c
ca
b
bc
anml
222
logloglog ++=++
××=
ab
c
ca
b
bc
a 222
log
1loglog222
222
==
cba
cba
10
10log
10
101log =×=
10log10log −= 011 =−=
Q.18. Prove that :
log loga
a b abb
− = ⋅
2 2( ) ( ) log log ( )
Ans. L.H.S. 22 )(log)(log ba −=
]log[log]log[log baba −+=
⋅=
b
aab log)(log
[∵ nmmn aaa logloglog += and ]logloglog nmn
maaa −=
Math Class IX 15 Question Bank
)(loglog abb
a⋅
= ...(i)
RHS )(loglog abb
a⋅
= ...(ii)
From (i) and (ii),
LHS = RHS
Q.19. (i) If a a+ = − −log ( 1) log (4 3) log3; find a
(ii) If log logy x− − =2 3 0, express x in terms of y.
(iii) Prove that: 2).log(13125log 1010 −=
Ans. (i) 3log)34(log)1(log −−=+ aa
⇒ 3
)34(log)1(log
−=+
aa
⇒ 3
)34()1(
−=+
aa ⇒ 34)1(3 −=+ aa ⇒ 3433 −=+ aa
⇒ 3343 −−=− aa ⇒ 6−=− a ⇒ 6=a
(ii) 03loglog2 =−− xy
⇒ 3loglog2 =− xy ⇒ 3loglog 2=− xy ⇒ 3log
2
=x
y
⇒ 1000loglog2
=x
y⇒ 1000
2
=x
y⇒ xy 10002
=
⇒ 21000 yx = ⇒ 1000
2yx =
(iii) )2log1(3125log 1010 −=
LHS 5log35log125log 103
1010 ===
RHS 5log3)5(log32
10log3)2log1(3 10101010 ==
=−=
∴ LHS = RHS
Math Class IX 16 Question Bank
Q.20. If log loga x b y= =2 3
, and loga b z− =2 3
3 2 6 , express y in terms of x and z.
Ans. 2 3log , loga x b y= = ⇒ zba log623
32=−
⇒ zyx log6log2log3 =− ⇒ 623 logloglog zyx =−
⇒ 6
2
3
loglog zy
x= ⇒ 6
2
3
zy
x= ⇒ 362
xzy = ⇒6
32
z
xy =
⇒ 2
1
6
3
=
y
xy ⇒
33
232
3
xy x z
z= = ÷ Hence,
3
32y x z= ÷
Q.21. If log loga b
a b−
= +1
log ( ),2 2
show that a b ab+ =2 2
6 .
Ans. log )log(log2
1
2ba
ba+=
−)(log
2
1ab= 2
1
)(log ab= ∴ 2
1
)(2
abba
=−
Squaring both sides, we get
212
2( )2
a bab
− =
⇒ ababba
=−+
4
222
⇒ ababba 4222
=−+
⇒ ababba 2422
+=+ ⇒ abba 622
=+
Q.22. If 23ab,ba22
=+ show that: logb)(loga2
1
5
balog +=
+
Ans. abba 2322
=+
⇒ abababba 223222
+=++ (Adding ab2 both sides, we get)
⇒ ababba 25222
=++ ⇒ abba 25)( 2=+
⇒ abba
=+
25
)( 2
⇒ abba
=
+2
5
Taking log both sides, we get
)(log5
log
2
abba
=
+ [∵ ]loglog mnm a
na =
⇒ )(log5
log2 abba
=+
⇒ )(log2
1
5log ab
ba=
+⇒ )log(log
2
1
5log ba
ba+=
+
Math Class IX 17 Question Bank
Q.23. Solve for x and y if 0x > and :0y > 2.2log2y
xloglogxy =+=
Ans. log log 2log 2 2x
xyy
= + =
(i) 10log2122log =×==xy 100log10log 2==
∴ 100=xy ...(i)
(ii) 22log2log =+y
x
100log2loglog2
=+y
x )2100log( =∵
100log2log2
=×y
x⇒ 100log
4log =
y
x⇒ 100
4=
y
x
⇒ yxyx 251004 =⇒= ...(ii)
From eqn. (i), we get
10025 =⋅ yy ⇒ 222 )2(410025 ±==⇒= yy
⇒ 2=y )0( >y∵
∴ 502
1001002 ==⇒=× xx
Hence, .2,50 == yx
Q.24. If m = log20 and n = log25, find the value of x, if: 2log ( 1) 2 .x m n+ = −
Ans. 25log,20log == nm
nmx −=+ 2)1(log2
⇒ 25log20log2)1(log2
−=+x ⇒ 25log)20(log)1(log22
−=+x
⇒ 25log400log)1(log2
−=+x ⇒ 2 400log ( 1) log
25x + =
∴ 22)4(16)1( ==+x ∴ 41 =+x ⇒ 314 =−=x ∴ 3=x
Math Class IX 18 Question Bank
Q.25. If
−−+−
45
7log2log3log16log2log7 log n= +1 , find the value of n.
Ans. nlog145
7log3log216log2log7log +=−−+−
⇒ nlog10log45
7log3log16log2log7log 2
+=−−+− )110log( =∵
⇒ nlog10log45
7log9log16log2log7log +=−−+−
⇒ )10(log)10(log792
45167log nn =×=
××
×× ∴
792
4516710
××
××=n
⇒ 479210
45167=
×××
××=n . Hence, 4=n
Q.26. Evaluate :
(i) 2
3
9 (ii) 3
2
8 (iii) 3
4
8−
(iv) 5
3
(243)
(v)
3
2
(27)
1
−
(vi) 5
2
32)(− (vii) 3
1
125
8−
(viii) 3
1
(0.001)−
Ans. (i)
3 3 3 3 32 1
22 2 2 2 19 (3 3) (3 ) 3 3× ×
= × = = = [ ( ) ]m n mna a=∵
2733333
=××==
(ii) 3
23
3
2
33
2
3
2
2)2()222(8×
==××= [ ( ) ]m n mna a=∵
422
==
(iii) 3
4
3
4
)222(8−−
××= 3
43
3
4
32)2(
−×−
== [ ( ) ]m n mn
a a=∵
16
1
2222
1
2
12
4
4=
×××===
− 1n
na
a
− =
∵
(iv) 5
3
5
3
)33333()243(−−
××××= 5
35
5
3
53)3(
−×−
==
3
3
3
1)3( ==−
27
1
333
1=
××=
1n
na
a
− =
∵
Math Class IX 19 Question Bank
(v) 3
2
3
2)27(
)27(
1=
−
1 n
na
a−
=
∵
3
2
33
2
)3()333( =××= 9333323
6
=×=== [ ( ) ]m n mn
a a=∵
(vi) 5
2
5
2
)]2()2()2()2()2[()32( −×−×−×−×−=−
5
25
5
2
5)2(])2[(
×
−=−= 4)2()2()2(2
=−×−=−= [ ( ) ]m n mn
a a=∵
(vii) 3
1
3
1
555
222
125
8−−
××
××=
2
5
5
2
5
2
5
213
3
13
13
=
=
=
=
−×−−
(viii) 3
1
3
1
)1.01.01.0()001.0(
−−
××=
101
101
10
1
1
1.0
1)1.0()1.0(
13
13
=×
=====−
−×
Q.27. Evaluate the following :
(i) 2
1
03
22
16
9583
4
1−−
+××−
(ii) 03
2
2
1
3(27)(0.01)4
1×−+
−
(iii)
÷
×
−
−− 3
2
3
4
3
2
5
9
25
16
81 (iv)
−
−
12 2
23(64) × 2 ÷ 70
(v) 3
2
2
3
4
3
216
343
9
49
81
16
÷
×
−
(vi)
−
−
2 03
1 3
4
64 1 25÷ +
125 64256
625
Math Class IX 20 Question Bank
(vii)
3
1
2
3
1
2
1
5
2
(64)(2)
(8)(4)(32)
−−
−
÷
×× (viii) 3
1
3
1
)532()532( +−
Ans. (i)
122203
1 93 8 5
4 16
− −
− × × +
2
1
03
22
4
3
4
35)222(3
2
1
2
1−−
×+××××−
×=
−×−×
+××−
=
2
12
03
2
3)2(2
4
35)2(3
2
1 ])([ mnnm aa =∵
12
4
4
3123
2
1−−
+××−
= ]1[
0=a∵
3
4143)2( 4
+××−=3
412)2222( +−×××=
=
−
m
m
aa
1∵
3
44
3
41216 +=+−=
3
15
3
16
3
412==
+=
(ii) 03
2
2
1
3)27()01.0(4
1×−+
−
03
2
2
12
12
3)333()1.01.0(2
1×××−×+
=
−
03
2
32
12
2
12
3)3()1.0(2
1×−+
=
−××
0211
3)3()1.0(2
1×−+
=
− ])([ mnnm aa =∵
191.0
1
2
1×−+=
==
−1and
1 0a
aa
m
m∵
2
119
2
1109
1
10
2
1=−=−+=
Math Class IX 21 Question Bank
(iii)
÷
×
−
−−3
2
3
4
3
2
5
9
25
16
81
÷
×
××
×××
×××=
−−−
32
3
4
3
2
5
33
55
2222
3333
÷
×
=
−×
−× 3
2
32
4
34
5
2
3
5
2
3
3 3 33 5 2
2 3 5
− − = × ÷
÷
×
=
333
5
2
5
3
3
2
××÷
×××
××
××=
5
2
5
2
5
2
5
3
5
3
5
3
333
222
18
27
27
8
8
125
125
27
27
8=×=
××=
(iv) 2
1
023
2
72)64(
−
−
÷×
2
1
23
2
12
1)444(
−
÷×××=
2
1
3
2
31
22
1)4(
−
÷
××=
2
1
3
23
14
14
−
×
÷×=2
1
2
1
21
4
1161
4
14
−−
××=
÷×=
2
1
2
1
)22(
1
4
1)4(
2
12
2
1
2
12
1
==
×
===
×
−
Math Class IX 22 Question Bank
(v) 3
2
2
3
4
3
216
343
9
49
81
16
÷
×
−
3
2
2
3
4
3
666
777
33
77
3333
2222
××
××÷
×
××
×××
×××=
−
3 3 24 2 34 2 32 7 7
3 3 6
−
= × ÷
3
23
2
32
4
34
6
7
3
7
3
2××
−×
÷
×
=
233
6
7
3
7
3
2
÷
×
=
−
66
77
333
777
2
33
×
×÷
××
×××
=
36
49
27
343
222
333÷×
××
××=
49
36
27
343
8
27××=
2
131
2
63==
(vi)
0
3
4
1
3
2
64
25
625
256
1
125
64
+
÷
−
0
3 3
2
4
1
3
2
4
5
5555
4444
1
555
444
+
×××
×××
÷
××
××=
−0
4
14
3
23
4
5
5
4
1
5
4
+
÷
=
−
1
5
4
1
5
4
4
14
3
23
+
÷
=
×
−×
]1[ 0=a∵
Math Class IX 23 Question Bank
14
5
4
51
5
4
1
5
42
1
2
+÷
=+
÷
=
−
=
−
m
m
aa
1∵
4
12
4
91
4
51
5
4
16
25==+=+×=
(vii)
3
1
2
3
1
2
1
5
2
)64()2(
)8()4()32(−
−
−
÷
××
3
1
2
3
1
2
1
5
2
)444()2(
)222()22()22222(−
−
−
××÷
×××××××××=
3
1
32
3
1
32
1
25
2
5
)4()2(
)2()2()2(−
−
−
÷
××=
−×
−
×
−××
÷
××=
3
13
2
3
13
2
12
5
25
)4()2(
)2()2()2(
122
112
12
112
)2(2
2
42
222−−
+−
−−
−
÷
=
÷
××= 4
1
22
22
222
2
=×
=
÷
=−−
(viii) 3
1
3 )532()532( +−
1
])([ mmm abba =∵
3
1
)]532)(532[( +−= 3
1
22])5()32[( −=
3
1
3
1
)27()532( =−= 3)3()333( 3
1
33
1
==××=
Q.28. Simplify :
(i) n n
n n
+
− −
×
×
1
1 1
3 9
3 9 (ii)
n n
n n n n
+ +
− + −÷
1 1
( 1) 1 1
3 9
3 (3 )
(iii) a b a b− − −
+ ⋅ +1 1 1
( ) ( ) (iv) n n
n n
+ +− ×
× − ×
3 1
2
5 6 5
9 5 5 2
Ans. (i) 121
12
11
1
)3(3
)3(3
93
93−−
+
−−
+
×
×=
×
×
nn
nn
nn
nn 2 2
1 2 2
3 3
3 3
n n
n n
+
− −
×=
×
33
23
221
22
3
3
3
3−
+
−+−
++
==n
n
nn
nn
5332333 ==
+−+ nn
Math Class IX 24 Question Bank
(ii) 11
1
)1(
1
)3(
9
3
3−+
+
−
+
÷nn
n
nn
n
1
)1)(1(1
1
11
)1(
1
)33(
3
3
3
9
)3(
3
32 +
−+
−
+
+
−+
−
+
×
×=×=n
nn
nn
n
n
nn
nn
n
)1(2
11
12
)1()(1
3
3
3
3
)3(
3
3
32
2
22
2 +
−
−
+
+
−
−
+
×=×=n
n
nn
n
n
n
nn
n
)22()(11
22
1122
2
23
3
3
3
3 +−−−−++
+
−
−
+
=×=nnnnn
n
n
nn
n
9
1
33
1
3
133
2
22211 22
=×
====−−−+−−++ nnnnn
(iii) )()( 111 −−−+⋅+ baba
+⋅
+=
baba
11
)(
1
ab
ab
baab
ab
ba
)(
)(
111
)(
1 +⋅
+=
×+×⋅
+=
abab
ba
ba
1)(
)(
1=
+⋅
+=
(iv) 2
13
2559
565
×−×
×−++
nn
nn
)49(5
)65(5
4559
5655 21121
−
−=
×−×
×−⋅=
+++
n
n
nn
nn
19)5(5
)19(55
)5(5
)625(551
=⋅
=−⋅
=n
n
n
n
Q.29. If a b c= ⋅ ⋅1960 2 5 7 , calculate the value of a b c− −
⋅ ⋅2 7 5 .
Ans. cba7521960 ⋅⋅=
⇒ 1960752 =⋅⋅cba
⇒ 775222752 ×××××=⋅⋅cba
⇒ 213752752 ××=⋅⋅
cba ...(i)
Comparing powers of 2, 5 and 7 on both sides of eqn. (i), we get
2,1,3 === cba
Hence, value of:
213572572
−−−−⋅⋅=⋅⋅
cba
232352
7
5
17
2
1
×
=××=
200
7
258
7
55222
7=
×=
××××=
Solve for x : (Q.30 to Q.32)
Math Class IX 25 Question Bank
Q.30. Evaluate
xa b
b a
−
=
1 3
Ans.
x
a
b
b
a31−
= ⇒
13)31(2
1−−−
=
=
xx
b
a
b
a
b
a
Comparing both sides, we get
2
31
2
13
2
113 =+=⇒=− xx
Hence, 3 1 1
2 3 2x = × =
Q.31. Evaluate
x−
=
1
32 27
3 8
Ans.
3
3
31
3
2
3
)2(
)3(
8
27
3
2
===
−x
⇒3
3
1
3
2
3
2−
−
=
x
Comparing both sides, we get
8199133
1−=+−=⇒−=−⇒−=
−xx
x. Hence, 8−=x
Q.32. Evaluate
x−−
=
1 13 27
5 125
Ans.
11
125
27
5
3−−
=
x
⇒
131
3
32
1
5
3
5
3
5
3−−−
=
=
x
⇒
32
1
5
3
5
3−
−
=
x
Comparing both sides, we get
6132
1−=−⇒−=
−x
x⇒ 516 −=+−=x . Hence, 5−=x
Math Class IX 26 Question Bank
Q.33. Solve for x:
(i) x x− +− =
2 14 2 0 (ii) x x
=2
3 : 3 9 : 1
(iii) x x x2 +18× 2 + 4× 2 = 1 + 2 (iv) x x
−2 +2 3
2 + 2 4× 2 = 0
(v) x x– 3 +14( 3) = ( 3)
Ans. (i) 02412
=−+− xx ⇒ 12
24+−
=xx
⇒ 12
24+−
=xx ⇒ 12 2)22( +−
=×xx ⇒ 1)2(2
22+−
=xx ⇒ 142
22+−
=xx
⇒ 2 4 1x x− = + ⇒ 2 1 4x x− = + ⇒ 5x =
(ii) 1:93:32
=xx
⇒ 931
9
3
3 22
=⇒=− xx
x
x
=
−nm
n
m
aa
a
⇒ 3332
×=− xx ⇒ 2
332
=− xx ⇒ 2
2=− xx
⇒ 022
=−− xx ⇒ 0222
=−+− xxx ⇒ 0)2(1)2( =−+− xxx
⇒ 0)2)(1( =−+ xx . If 0)2( =−x then 2=x . If 0)1( =+x then 1−=x
(iii) xxx212428
12+=×+×
+ ⇒ xxx 21224)2(8 12+=××+×
⇒ xxx2128)2(8
2+=×+× ⇒ 01228)2(8
2=−−×+×
xxx
⇒ 01)18(2)2(8 2=−−+×
xx ⇒ 0127)2(82
=−×+xx
]2 Putting[ yx
=
⇒ 017)(8 2=−×+× yy ⇒ 0178 2
=−+ yy
⇒ 0188 2=−−+ yyy ⇒ 0)1(1)1(8 =+−+ yyy
⇒ 0)18)(1( =−+ yy ⇒ 0)1( =+y and 018 =−y
⇒ 012 =+x and 0128 =−×
x
⇒ 12 −=x
and 128 =×x ⇒ 12 −=
x and
8
12 =
x
⇒ 12 −=x
and 3
2
12 =
x
⇒ 12 −=x
and 3223
−=⇒=−
xx
Math Class IX 27 Question Bank
(iv) 02422 322=×−+
+xx
⇒ 2 2(2 ) 2 2 4 2 2 2 0x x+ ⋅ − × × × = ⇒ 2(2 ) 2 4 32 0·x x
+ − =
⇒ 03224)2( 2=−×+
xx ⇒ 03242=−+ yy [Putting ]2 y
x=
⇒ 032482
=−−+ yyy ⇒ 0)8(4)8( =+−+ yyy
⇒ 0)4)(8( =−+ yy ⇒ 08 =+y and 04 =−y
⇒ 8−=y and 4=y ⇒ 82 −=x and 2
22 =x
⇒ 2=x
(v) 143)3()3(
+−=
xx
⇒
1
4
13
2
1
33
+−
=
xx
⇒1
4
13
2
1
33+×−×
=
xx
⇒ )1(4
1)3(
2
1+×=−× xx ⇒
4
)1(
2
)3( +=
− xx
⇒ )1(2)3(4 +=− xx ⇒ 22124 +=− xx
⇒ 12224 +=− xx ⇒ 72
14142 =⇒=⇒= xxx
Q.34. Solve for x if :(i)x+
=
12
53 32( 4) ( 8) (ii) x xp p q q p q
− −× = × ≠
1 2 1 2; .
Ans. (i) 532
12
3 )8()4( =
+x
⇒ 532
12
3 )222()22( ××=×
+x
⇒ 53 32
12
3 2)2()2( =
+x
⇒
2 1 32
53 2 32 (2 )x
+
= ⇒
2 12
1 53 22 (2 )x
+
=
⇒ 512
12
3
2
22×
+
=
x
⇒ 52
12
3
2
22 =
+x
⇒ 52
12
3
2=
+x ⇒ 5
3
2
1222
=
×+× x
⇒ 53
14=
+x⇒ 15143514 =+⇒×=+ xx
⇒ 1441154 =⇒−= xx ⇒2
13
2
7
4
14=⇒=⇒= xxx
Math Class IX 28 Question Bank
(ii) )(;2121 qpqqpp xx≠×=×
−− ⇒ 2
1
212
1
21qqpp
xx×=×
−−
⇒ 2
121
2
121 +−+−
=
xx
qp ][ nmnmaaa
+=×
But this is true only
If 02
121 =+− x ]1[
00== qp∵
⇒ 2
112 −−=− x ⇒
2
32
2
122
−=−⇒
−−=− xx
⇒ 22
3
2
32
×=⇒= xx ⇒
4
3=x
Q.35. Prove that :
(i)
a b b c c aa b c
b c a
x x x
x x x
+ + +
⋅ ⋅ =
1
(ii)
1 1 1a b cab bc ca
b c a
x x x
x x x
⋅ ⋅ =
1
(iii)
( ) ( ) ( )a b c b c a c a ba b c
b c a
x x x
x x x
+ − + − + −
⋅ ⋅ =
1
Ans. (i) LHS
ac
a
ccb
c
bba
b
a
x
x
x
x
x
x+++
⋅
⋅
=
))(())(())(( acaccbcbbabaxxx
+−+−+−⋅⋅=
222222 accbba
xxx−−−
⋅⋅=
====−+−+− 10222222
xxaccbba RHS
(ii) LHS ca
a
cbc
c
bab
b
a
x
x
x
x
x
x
111
⋅
⋅
=
caac
bccb
abba
xxx
1)(
1)(
1)( −−−
⋅⋅=
ca
ac
bc
cb
ab
ba
x
−+
−+
−
=
abc
acbcbabac
x
)()()( −+−+−
=
Math Class IX 29 Question Bank
====
−+−+−
10
xx abc
abbccaabbcca
RHS
(iii) LHS
)()()( bac
a
cacb
c
bcba
b
a
x
x
x
x
x
x−+−+−+
⋅
⋅
=
))(())(())(( bacacacbcbcbabaxxx
−+−−+−−+−⋅⋅=
abbcaccaabcbbcacbax
+−−++−−++−−=
222222
=== 10
x RHS
Q.36. Simplify :
(i)
2 2 2 2 2 2a ab b b bc c c ca aa b c
b c a
x x x
x x x
+ + + + + +
⋅ ⋅
(ii)
2 2 2 2 2 2a ab b b bc c c ca aa b c
b c a
x x x
x x x
− + − + − +
− − −
× ×
Ans. (i)
222222 acac
a
ccbcb
c
bbaba
b
a
x
x
x
x
x
x++++++
⋅
⋅
222222
)()()(acacaccbcbcbbababa
xxx++−++−++−
××=
))(())(())(( 222222 acacaccbcbcbbababaxxx
++−++−++−××=
333333 accbba
xxx−−−
××=
10333333
===−+−+−
xxaccbba
(ii)
222222acac
a
ccbcb
c
bbaba
b
a
x
x
x
x
x
x+−
−
+−
−
+−
−
×
×
222222
)()()(acacaccbcbcbbababa
xxx+−++−++−+
××=
))(())(())(( 222222 acacaccbcbcbbababaxxx
+−++−++−+××=
333333 accbba
xxx+++
××=333333 accbba
x+++++
=)(2 333 cba
x++
=
Math Class IX 30 Question Bank
Q.37. If ,x ya b b c= = and z
c a= , prove that : xyz = 1.
Ans. We are given that cbbayx
== , and ac z=
bax
=
⇒ yzxyzba = (Raising to the power yz both sides)
⇒ zzyxyzcba == )( )( cb
y=∵
⇒ aaxyz
= )( ac z=∵
⇒ 1aa
xyz=
Comparing both sides, we get
⇒ 1=xyz or zca =
zyb )(=
yzb= )(
ybc =∵
yzxa )(=
xyza= )( xab =∵
Comparing both sides, we get 1=xyz
Q.38. If x y za b c= = and 2 ,b ac= prove that :
2.
xzy
x z=
+
Ans. Let kcbazyx
===
∴ yx kbka
11
, == and zkc
1
=
Now acb =2
⇒ zxy kkk
112
1
×=
⇒ zxy kk
112+
=
Comparing both sides, we get
xzzxyxz
zx
yzxy2)(
2112=+⇒
+=⇒+=
Hence, zx
xyy
+=
2.
Math Class IX 31 Question Bank
Q.39. If x y z= =2 3 12 , show that :
z y x= +
1 1 2.
Ans. Let ,1232 kzyx
=== then
xxkk
1
22 =⇒= , yykk
1
33 =⇒=
zzkk
1
1212 =⇒= . Now 3232212 2×=××=
⇒ yxz kkk
1
2
11
)( ×= ⇒ yxz kkk
121
×= ⇒ yxz kk
121+
=
Comparing both sides, we get
yxz
121+=
Q.40. If x y z−= =2 3 6 show that :
x y z+ + =
1 1 10.
Ans. Let ,632 kzyx
===− then zyx kkk
111
6,3,2−
===
Now 632 =×
⇒ zyx kkk
111 −
=× ⇒ zyxkk
111−+
=
Comparing both sides, we get
zyx
111−=+ ⇒ 0
111=++
zyx