ML Aggarwal Solution Class 9 Chapter 9 Logarithms Exercise 9.2
Exercise 9.2
Question 1
Simplify the following :
=log32−log3log33
=2log3−log33log3 (Power Law)
=log33log3=13
Question 2
Evaluate the following:
Sol :
=log((10)1÷(10)13)
=log(101−13)
=log(1023)
=23log10=23(1)=23
(ii) 2+12log(10−3)
Sol :
=2+12×(−3)log10
=2−32log10=2−32(1)
=2−32=4−32=12
(iii) log5+log8−12log4
Sol :
=log(5)2+log8−12log(2)2
=log25+log8−12×2log2
=log25+log8−log2=log(25×82)
=log(25×41)
=log (100)
=log (10)2
=2 log 10=2(1)=2
(iv) 2log103+3log10−2−13log5−3+12log4
Sol :
=2×3log10+3(−2)log10−13(−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+log5−12log36−log130
Sol :
=log(2)2+log5−12log(6)2−log(130)
=log4+log5−12×2log6−log130
=log4+log5−12×2log6−log130
=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+3log2−12log36−2log10
Sol :
=log (5)2+log 3+log (2)3−12log(6)2−2log10
=log 25+log 3+log 8−12×2log6−2log10
=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)
(viii) 2log105+log108−12log104
Sol :
=log10(5)2+log108−log10(4)12
=log1025+log108−log10(2)2×12
=log1025+log108−log102
=log10(25×82)=log10(25×4)
=log10100=log10(10)2
=2log1010=2(1)=2
Question 3
=12log9+2log3−log6+log2−2
=log912+log32−log6+log2−log100
=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 9L.H.S=log10 4 ÷ log10 2
L.H.S=log10 25 + log10 4
=log10 25×4
=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 log10√yx2z3 in terms of a,b,c
=(10)2a
y=(10000)b=[(10)4]b
=(10)4b
z=(10)c=(10)c
Now, log10√yx2z3
=(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 :
Sol :
(i) Given that,
⇒a=log10 x
⇒(10)a=x
⇒x=(10)a
(ii) log105√x2
=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=2log√52
Find the value of
(i) a+b+c
(ii) 5a+b+c
Sol :
Given that :
a=log23,b=log35,c=2log√52
(i) a+b+c=log23+log35+2log√52
=(log2−log3)+(log3−log5)+2log(52)12
=log2−log3+log3−log5+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=2log√32 Find the value of
(i) x+y-z
(ii) 3x+y-z
Sol :
x=log35,y=log54,z=2log√32
∴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
(i) x-y-z
= log10 12-log4 2×log10 9- log10 0.4
= log10 (3×4)-log4 41/2×log10 32- log10 410
==log103+log104−12log44×2log103−(log104−log1010)
=log103+log104−12×1×2log103−log104+1
=log103+log104−log103−log104+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.
⇒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.
⇒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)2−log(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.
Question 13
Express log10 2+1 in the from log10 x.
Question 14
Question 15
Question 16
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)2−log5
⇒log x=log 9-log 5
⇒logx=log(95)
∴x=95
(ii) log3x−log32=1
Sol :
⇒log3x−log32=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×log3log√3=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
⇒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)log1−log3×log5
⇒logx=log(3)20−log3×log5 [log 1=0]
⇒logx=2log3−log3×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)2log1−log3×log2
⇒logy2=2log30−log3×log2 [log 1=0]
⇒logy2=2log3−log3×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 log22⋅log3=log3⋅log22
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
⇒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(x2−1)=log24
⇒x2−1=24
⇒x2=24+1
⇒x2=25
⇒x2=(5)2
∴x2=5
(iii) log (10x + 5) – log (x – 4) = 2
⇒log(10x+5)(x−4)=2(log10) [∴log 10=1]
⇒log10x+5x−4=log(10)2
⇒log(10x+5x−4)=log100
⇒10x+5(x−4)=100
⇒10x+5=100(x-4)
⇒10x+5=100x-400
⇒10x-100x=-400-5
⇒-90x=-405
⇒x=−405−90
⇒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
⇒log4y−3=log(2y+1)3
⇒4y−3=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(x−2)=log103+3log104
⇒log10(x+2)(x−2)=log103+log10(4)3
⇒log10(x2−22)=log103+log10(4×4×4)
⇒log10(x2−4)=log103+log1064
⇒log10(x2−4)=log103×64
⇒x2−4=3×64
⇒x2−4=192
⇒x2=192+4
⇒x2=196
⇒x2=(14)2
∴x=14
Hence proved
(vii) log(3x + 2) + log(3x – 2) = 5 log 2.
Question 23
Solve for x :
Sol :
⇒log3(x+1)−1=3+log3(x−1)
⇒log3(x+1)−3log(x−1)=3+1
⇒log3x+1x−1=4=4×1=4log33 (∵logaa=1)
⇒log3x+1x−1=log334=log381
∴x+1x−1=811
⇒81x-81=x+1
⇒81x-x=1+81
⇒80x=82
∴x=8280=4140=1140
Question 24
Solve for x:5logx+3logx=3logx+1−5logx−1
Sol :
⇒5logx+3logx=3logx+1−5logx−1
⇒5logx+3logx=3logx⋅31−5logx⋅5−1
⇒5logx+3logx=3.3logx−15⋅5logx
⇒5logx+15⋅5logx=3.3logx−3logx
⇒(1+15)(5logx)=(3−1)(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(x−y2)=12(logx+logy), prove that x2+y2=6xy
Sol :
⇒log(x−y2)≐12(logx+logy)
⇒log(x−y2)=12logxy [∵log m+log n=log mn]
⇒log(x−y2)=log(xy)12
⇒⇒xy2=(xy)12
Question 26
If x2+y2=23xy, Prove that logx+y5=12(logx+logy)
Sol :
⇒Given x2+y2=23xy
⇒x2+y2=25xy−2xy
⇒(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)=2p−q
Question 28
Show that:
(i) 1log242+1log342+1log742=1
Sol :(ii) 1log836+1log936+1log1836=2
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