ML Aggarwal Solution Class 9 Chapter 9 Logarithms Exercise 9.2

 Exercise 9.2

Question 1

Simplify the following :

(i) log a3-log a2
Sol :
⇒log a3-log a2
=log(a3a2) (Quotient Law)
=log a

(ii) log a3÷log a2
Sol :
=3log a÷2log a (Power Law)
=3loga2loga=32

(iii) log4log2
Sol :
=log(2×2)log2=log2log2
=2log2log2 (Power Law)
=2(1)=2

(iv)log8log9log27
Sol :
=log23log32log33
=(3log2)(2log3)(3log3) (Power Law)
=(log2)(2)1
=2 log 2=log 22
=log 4

(v) log27log3
Sol :
=log(3×3×3)log(3)12
=log(3)3log(3)12
=3log312log3 (Power Law)
=3×21(log3log3)
=6(1)=6


(vi) log9log3log27
Sol :
=log(3×3)log3log(3×3×3)

=log32log3log33

=2log3log33log3 (Power Law)

=log33log3=13


Question 2

Evaluate the following:

(i) log(10÷310)

Sol :

=log((10)1÷(10)13)

=log(10113)

=log(1023)

=23log10=23(1)=23


(ii) 2+12log(103)

Sol :

=2+12×(3)log10

=232log10=232(1)

=232=432=12


(iii) log5+log812log4

Sol :

=log(5)2+log812log(2)2

=log25+log812×2log2

=log25+log8log2=log(25×82)

=log(25×41)

=log (100)

=log (10)2

=2 log 10=2(1)=2


(iv) 2log103+3log10213log53+12log4

Sol :

=2×3log10+3(2)log1013(3)log5+12

=log (2)2

=6 log 10-6 log 10+33log5+12×2log2

=0+1 log 5 +log 2=log 5+log 2

=log (5×2)

=log(10)=1


(v) 2log2+log512log36log130

Sol :

=log(2)2+log512log(6)2log(130)

=log4+log512×2log6log130

=log4+log512×2log6log130

=log 4+log 5-log 6-(log 1-log 30)

=log 4+log 5-log 6-log 1+log 30

=(log 4+log 5+log 30)-(log 6+log 1) 

=log (4×5×30)-log (6×1)

=log4×5×306×1=log4×5×51×1

=log 100

=log (10)2

=2 log 10

=2(1)=2


(vi) 2log5+log3+3log212log362log10

Sol :

=log (5)2+log 3+log (2)312log(6)22log10
=log 25+log 3+log 812×2log62log10

=log 25+log 3+log 8-log 6-log (10)2

=log (25×3×8)-log 6-log 100

=log(25×3×86×100)=log(1×3×86×4)

=log(2424)

=log 1=0


(vii) log2+16log1615+12log2524+7log8180

Sol :

=log 2+16(log 16-log 15)+12(log 25-log 24)+7(log 81-log 80)

=log 2+16[log (2)4-log (3×5)]+12[log (5)2-log(3×2×2×2)+7[log(3×3×3×3)-log (2)4×5]

=log 2+16[4 log 2-(log 3+log 5)]+12[2 log 5-log(3×23)]+7[log(3)4-(log 4+log 5)]

=log 2+16[4 log 4- log 3 - log 5]+12[2 log 5-log (3×23)]+7[log 34-(log 4+ log 5)]

=log 2+16[4 log 4-log 3-log 5]+12[2 logs 5 -(log 3-log 23)]+7[4 log 3-log 4-log 5]

=log 2+ 64 log 2-16 log 3-16 log 5+24 log 5-12 log 3-12 log 23+28 log 3-7 log 24-7 log 5

=log 2+ 64 log 2-16 log 3-16 log 5+24 log 5-12 log 3-36 log 2+28 log 3-28 log 2-7 log 5

=(log 2+ 64 log 2-36 log 2-28 log 2)+(-16 log 3-12 log 3+28 log 3)+(-16 log 5+24 log 5-7 log 5)+28 log 3)+(-16 log 5+24 log 5-7 log 5)

=(65 log 2-64 log 2)+(-28 log 3+28 log 3)+(-23 log 5+24 log 5)
=log 2+0+log 5=log 2+log 5
=log(2×5)
=log 10 =1

(viii) 2log105+log10812log104

Sol :

=log10(5)2+log108log10(4)12

=log1025+log108log10(2)2×12

=log1025+log108log102

=log10(25×82)=log10(25×4)

=log10100=log10(10)2

=2log1010=2(1)=2


Question 3

Express each of the following as a single logarithm:
(i) 2log312log16+log12
=2log312log(4)2+log12
=2log312×2log4+log12
=2log3log4+log12
=log(3)2log4+log12
=log 9-log 4+log 12
=log9×124=log(9×31)
=log 27


(ii) 2log105log102+3log104+1
=log10(5)2log102+log10(4)3+log1010 (log1010=1)
=log1025log102+log1064+log1010
=log10(25×64×10)log102
=log10(16000)log102
=log10(160002)=log108000


(iii) 12log36+2log8log1.5
=log(36)12+log(8)2log1.5
=log(6)2×12+log64log(1510)
=log 6+log 64-(log 15-log 10)
=log 6×64-log 15+log 10
=log (6×64×10)-log 15
=log(60×6415)
=log (4×64)
=log 256


(iv) 12log252log3+1
=log(25)12log(3)2+log10 (∵ log 10=1)
=log(5)2×12log9+log10
=log 5-log 9+log 10
=log (5×10)-log 9
=log5×109=log509


(v) 12log9+2log3log6+log22

=12log9+2log3log6+log22

=log912+log32log6+log2log100

=log 3+log 9-log 6+log 2-log 100

=log3×9×26×100

=log9100


Question 4

Prove the following :

(i) log10 4 ÷ log10 2 =  log3 9
L.H.S=log10 4 ÷ log10 2 
=2log102log102
=2(1)=2

R.H.S=log3 9
=log(3)2 
=2(1)=2

Hence, proved L.H.S=R.H.S

(ii) log10 25 + log10 4 = log5 25
L.H.S=log10 25 + log10 4 
=log10 25×4 
=log10 100
=log10 102

=2 log10 10

=2×1=2 (∵log10 a=1)

Hence, L.H.S=R.H.S


Question 5

If x = 100)a , y = (10000)b and z = (10)c, express log10yx2z3 in terms of a,b,c

Sol :
Given that x=(100)a=[(10)2]a

=(10)2a


y=(10000)b=[(10)4]b

=(10)4b


z=(10)c=(10)c

Now, log10yx2z3

=(log 10+log √y)-(log x2+logz3)

=(1+log(y)12)(log(x)2+log(z)3) [∵ log 10=1]

=(1+12logy)(2logx+3logz)

Substituting the value of x, y and z, we get

=(1+12log(10)4b)(2log(10)2a+3log(10)c)

=(1+12×4 blog10)(2×2alog10+3×clog10)

=(1+12×4 b×1)(2×2a×1+3×c×1) [∵log 10=1]

=(1+2b)-(4a+3c)

=1+2b-4a-3c

=1-4a+2b-3c


Question 6

If a = log10 x, find the following in terms of a :

(i) x
(ii) log105x2
(iii) log105x

Sol :

(i) Given that,

⇒a=log10 x

⇒(10)a=x

⇒x=(10)a


(ii) log105x2

=log10(x2)15

=log10(x)25

=25log10x

=25(a)=25a


(iii) x=(10)a=log10 5x

=log10 5(10)a

=log10 5+log10 10

=log10 5+a(1)

=a+log10 5


Question 7

If a=log23,b=log35 and c=2log52

Find the value of

(i) a+b+c

(ii) 5a+b+c

Sol :

Given that :

a=log23,b=log35,c=2log52

(i) a+b+c=log23+log35+2log52

=(log2log3)+(log3log5)+2log(52)12

=log2log3+log3log5+2×12log(52)

=log 2-log 3+log 3-log 5+log \frac{5}{2}$

=log 2+(-log 3+log 3)-log 5+log 5-log 2$

=(log 2-log 2)+0+(log 5-log 5)

=0+0+0=0


(ii) 5a+b+c=50

=1


Question 8

If x=log35,y=log54 and z=2log32 Find the value of 

(i) x+y-z 

(ii) 3x+y-z

Sol :

x=log35,y=log54,z=2log32

∴x=log 3-log 5

y=log 5-log 4

z=log(32)2=log34

=log 3-log 4

(i) Now, x+y-z=log 3-log 5+log 5-log -log 3-log 4

=0

(ii) 3x+y-3

=30=1


Question 9

If x = log1012, y = log4 2×log10 9 and z = log10 0.4, find the values of

(i) x-y-z

(ii) 7x-y-z

Sol :
x = log10 12, y = log4 2×log10 9 and z = log10 0.4

(i) x-y-z

= log10 12-log4 2×log10 9- log10 0.4

= log10 (3×4)-log4 41/2×log10 32- log10 410

==log103+log10412log44×2log103(log104log1010)

=log103+log10412×1×2log103log104+1

=log103+log104log103log104+1

=1


(ii) 7x-y-z=71=0


Question 10

If log V + log 3 = log π + log 4 + 3 log r, find V in terns of other quantities.

Sol :
logV+log3=logπ+log4+log(r)3
log(V×3)=log(π×4×r3)

log(3 V)=log4πr3

3 V=4πr3

V=43πr3


Question 11

Given 3 (log 5 – log 3) – (log 5-2 log 6) = 2 – log n , find n.

Sol :
Given that 3(log 5-log 3)-(log 5-2 log 6)= 2-log n

⇒3 log 5-3  log 3- log 5+2 log 6=2- log n 

⇒2 log 5-3 log 3+2 log 6=2-log n

log(5)2log(3)3+log(6)2=2(1)logn

⇒log 25-log 27+log 36=2 log 10- log n  [∵log 10=1]

⇒log n=2 log 10-log 25+log 27-log 36

⇒log n=log (10)2-log 25+log 27-log 36

⇒log n=log 100-log 25+log 27-log 36

⇒log n=(log 100+log 27)-(log 25+log 36)

⇒log n=log (100×27)-log (25×36)

logn=log(100×2725×36)

logn=log(4×271×36)

logn=log(1×271×9)

⇒log n=log 3

⇒n=3


Question 12

Given that log10 y + 2 log10 x= 2, express y in terms of x.

Sol :
⇒log10 y + 2 log10 x= 2
⇒log10 y + log10 x2= 2
⇒log10 yx2= 2
⇒log10 yx2= 2 log10 10
⇒log10 (yx2)= 2 log10 (10)2
⇒yx2=(10)2
⇒yx2=100
y=100x2


Question 13

Express log10 2+1 in the from log10 x.

Sol :
⇒log10 2+1=log10 2+log10 10 (∵ log10 10=1)
⇒log10 2×10=log10 20

Question 14

If a2=log10x,b3=log10y and a22b23=log10z express z in terms of x and y
Sol :
Given that
a2=log10x,b3=log10y
⇒We have a22b23=log10z
12(log10x)13(log10y)=log102
log10(x)12log10(y)13=log102
log10xlog103y=log10z
log10x3y=log10z
x3y=z
Hence , z=x3y

Question 15

Given that log m = x + y and log n = x-y, express the value of log m2n in terms of x and y.
Sol :
Given that log m=x+y and log n=x-y
⇒log mn=log m+log n
=2 log m+log n
=2(x+y)+x-y
=2x+2y+x-y
=3x+y

Question 16

Given that log x=m+n and log y=m-n, express the value of log(10xy2) in terms of m and n
Sol :
Given that log n=m+n and log y=m-n
Then log(10xy2)=log10xlogy2
=log10+logx2logy
=1+log x-2 log y
=1+(m+n)-2(m-n)
=1+m+n-2m+2n
=1-m+3n

Question 17

If logx2=logy3, find the value of y4x6

Sol :

logx2=logy3

⇒3 log x=2log y

logx3=logy2

x3=y2

Squaring both sides, we get

x6=y4

y4=x6

y4x6=1


Question 18

Solve for x:

(i) log x+log 5=2 log 3

Sol :

⇒log x+log 5=2 log 3

⇒log x=2 log 3- log 5 

logx=log(3)2log5

⇒log x=log 9-log 5

logx=log(95)

x=95


(ii) log3xlog32=1

Sol :

log3xlog32=1 

log3x=log32+1

logyx=log32+log33 (log33=1)

log3x=log3(2×3)

log3x=log36

∴x=6


(iii) x=log125log25

Sol :

x=log(5)3log(5)2x=log(5)3log(5)2

x=3log52log5=32

x=32


(iv) log8log2×log3log3=2logx

Sol :

log(2)3log2×log3log(3)12=2logx

3log2log2×log312log3=2logx

3(1)×1(12)=2logx

3×21=2logx

⇒2 log x=6

logx=62

⇒log x=3

x=(10)3

⇒x=1000


Question 19

Given 2log10x+1=log10250, find

(i) x

(ii) log102x

Sol :
(i) 2log10x+1=log10250

log10x2+1=log10250  [logamn=nlogm]

log10x2×10=log10250 [log1010=1]

log10x2×log1010=log10250

x2×10=250

x2=25010

x2=25

(x)2=(5)2

∴x=5


(ii) x=5 (proved in (i) above)

log102x=log102×5 [Putting x=5]

=log1010=1 [log1010=1]


Question 20

If logxlog5=logy2log2=log9log13, find x and y.

Sol :

logxlog5=logy2log2=log9log13

Taking first and third terms,

logxlog5=log9log13

logx=log9log13×log5

logx=3log(3×3)log1log3×log5

logx=log(3)20log3×log5 [log 1=0]

logx=2log3log3×log5

logx=2log3log3×log5

⇒log x=-2(1)×log 5

⇒log x=-2log 5

⇒log x=log (5)-2

⇒x=(5)-2

x=1(5)2

x=125

taking second and third terms,

logy2log2=log9log(13)

logy2=log9log(13)×log2

logy2=log(3)2log1log3×log2

logy2=2log30log3×log2 [log 1=0]

logy2=2log3log3×log2

logy2=2log3log3×log2

logy2=2×log2

logy2=log(2)2

y2=(2)2

y=(2)2×12

y=(2)1

y=12


Question 21

Prove the following :

(i) 3log4=4log3

Sol :

3log4=4log3 is true

if log3log4=log4log3 (Taking log both sides)

if log 4.log 3= log 3.log 4

if log22log3=log3log22

if 2log 2×log 3=log 3×2log 2

if 2 log 2 log 3=2 log 2 log 3

Which is true

Hence proved


(ii) 27log2=8log3

Sol :
⇒ if log27log2=log8log3 (Taking log both sides)

⇒if log 2 log 27=log 3 log 8

⇒ if log2log33=log3log23

⇒if log 2.3 log 3=log 3.3 log 2

⇒if 3 log 2.log 3=3.log 2 log 3

Which is true

Hence proved


Question 22

Solve the following equations :

(i) log (2x + 3) = log 7

⇒2x+3=7

⇒2x=7-3

⇒2x=4

x=42

∴x=2


(ii) log (x +1) + log (x – 1) = log 24

⇒log(x+1)(x-1)=log 24

log(x21)=log24

x21=24

x2=24+1

x2=25

x2=(5)2

x2=5


(iii) log (10x + 5) – log (x – 4) = 2

log(10x+5)(x4)=2(log10) [∴log 10=1]

log10x+5x4=log(10)2

log(10x+5x4)=log100

10x+5(x4)=100

⇒10x+5=100(x-4)

⇒10x+5=100x-400

⇒10x-100x=-400-5

⇒-90x=-405

x=40590

x=40590=8118=92

∴x=4.5


(iv) log105+log10(5x+1)=log10(x+5)+1

log105×(5x+1)=log10(x+5)+log1010 [log1010=1]

log105(5x+1)=log10[10×(x+5)]

⇒5(5x+1)=10(x+5)

⇒25x+5=10x+5

⇒25x-10x=50-5

⇒15x=45

x=4515

∴x=3


(v) log (4y – 3) = log (2y + 1) – log3

log4y3=log(2y+1)3

4y3=2y+13

⇒3(4y-3)=2y+1

⇒12y-9=2y+1

⇒12y-2y=1+9

⇒10y=10

y=1010=1

∴y=1


(vi) log10(x+2)+log10(x2)=log103+3log104

log10(x+2)(x2)=log103+log10(4)3

log10(x222)=log103+log10(4×4×4)

log10(x24)=log103+log1064

log10(x24)=log103×64

x24=3×64

x24=192

x2=192+4

x2=196

x2=(14)2

∴x=14

Hence proved


(vii) log(3x + 2) + log(3x – 2) = 5 log 2.

log(3x+2)(3x2)=log25
log(9x24)=log32
Comparing both sides
9x24=32
9x2=32+4=36
x2=369=4=(±2)2
∴x=±2
x=2

Question 23

Solve for x :

log3(x+1)1=3+log3(x1)

Sol :

log3(x+1)1=3+log3(x1)

log3(x+1)3log(x1)=3+1

log3x+1x1=4=4×1=4log33 (logaa=1)

log3x+1x1=log334=log381

x+1x1=811

⇒81x-81=x+1

⇒81x-x=1+81

⇒80x=82

x=8280=4140=1140


Question 24

Solve for x:5logx+3logx=3logx+15logx1

Sol :

5logx+3logx=3logx+15logx1

5logx+3logx=3logx315logx51

5logx+3logx=3.3logx155logx

5logx+155logx=3.3logx3logx

(1+15)(5logx)=(31)(3logx)

65(5logx)=2×3logx

5logx3logx=2×56=(53)1

(53)logx=(53)1

Comparing, we get

⇒log x=1=log 10

∴x=10


Question 25

If log(xy2)=12(logx+logy), prove that x2+y2=6xy

Sol :

log(xy2)12(logx+logy)

log(xy2)=12logxy [∵log m+log n=log mn]

log(xy2)=log(xy)12

xy2=(xy)12

⇒Squaring both sides, we get
(xy2)2=[(xy)12]2
(xy)24=(xy)12×2
(xy)2=4×xy
x2+y22xy=4xy
[(AB)2=A2+B22AB]
x2+y2=4xy+2xy
x2+y2=6xy

Question 26

If x2+y2=23xy, Prove that logx+y5=12(logx+logy)

Sol :

⇒Given x2+y2=23xy

x2+y2=25xy2xy

(x)2+(y)2+2×x×y=25xy

(x+y)2=25xy

(x+y)225=xy

Taking log on both sides, we get

log(x+y)225=logxy

log(x+y5)2=logx+logy

2logx+y5=logx+logy

logx+y5

=12(logx+logy) Proved


Question 27

If p=log1020 and q=log1025, find the value of x if 2log10(x+1)=2pq

Sol :
Given that p=log1020 and q=log1025
Then, 2log10(x+1)=2pq
Substituting the value of p and q, we get
2log10(x+1)=2log1020log1025
2log10(x+1)=2log1020log10(5)2
2log10(x+1)=2log10202log105
2log10(x+1)=2(log1020log105)
log10(x+1)=2(log1020log105)2
log10(x+1)=log1020log105
log10(x+1)=log10(205)
log10(x+1)=log104
⇒x+1=4
⇒x=4-1
∴x=3

Question 28

Show that:

(i) 1log242+1log342+1log742=1

Sol :
1log242+1log342+1log742=1
L.H.S=1log242+1log342+1log742
{lognm=logmlogn}

=1log42log2+1log42log3+1log42log7
=log2log42+log3log42+log7log42
=log2+log3+log7log42=log(2×3×7)log42
{logm+logn+logp=logmnp}
=log42log42=1
=R.H.S

(ii) 1log836+1log936+1log1836=2

Sol :
L.H.S=1log836+1log936+1log1836
=1log36log8+1log36log9+1log36log18
=log8log36+log9log36+log18log36
=log8+log9+log18log36
=log(8×9×18)log36
=log(36)2log36
=2log36log36
=2=R.H.S

Question 29

Prove the following identities:
(i) 1logaabc+1logbabc+1logcabc=1
Sol :
1logaabc+1logbabc+1logcabc=1
⇒L.H.S. =1logaabc+1logbabc+1logcabc
=1logabcloga+1logabclogb+1logabclogc
{lognm=logmlogn}
=logalogabc+logblogabc+logclogabc
=loga+logb+logclogabc
=logabclogabc=1
=R.H.S
{logmnp=logm+logn+logp}


(ii) logbalogcblogdc=logda
Sol :
L.H.S=logba×logcb×logdc

=logalogb×logblogc×logclogd=logalogd

=logda
=R.H.S

Question 30

Given that logax=1α,logbx=1β,logcx=1γ
Find logabcx
Sol :
logax=1α,logbx=1β,logcx=1gamma
logax=1α
logxloga=1α
loga=αlogx
logbx=1β
logxlogb=1β
logb=βlogx
logcx=1γ
logxlogc=1γ
⇒ logc=γlogx

Now logabcx=logxlogabc
=logxloga+logb+logc
=logxαlogx+βlogx+γlogx
=logxlogx(α+β+γ)=1α+β+γ

Question 31

Solve for x
(i) log3x+log9x+log81x=74
Sol :
1logx3+1logx9+1logx81=74
1logx31+1logx32+1logx34=74
1logx3+12logx3+14logx3=74
1logx3[1+12+14]=74
logx3×74=74
logx3=74×47=1=log33
{logna=1}

Comparing, we get
∴x=3

(ii) log2x+log8x+log32x=2315
Sol : 
1logx2+1logx8+1logx32=2315
1logx21+1logx23+1logx25=2315
1logx2+13logx2+15logx2=2315
1logx2[1+13+15]=2315
2315[logx2]=2315
logx2=2315×1523=1=log22 {logaa=1}

Comparing , we get
x=2

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2