ML Aggarwal Solutions Class 9 Math 9th Chapter Logarithms Exercise 9.2
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Ninth Chapter Logarithms Exercise 9.2. APC Solution Class 9 Exercise 9.2.
Exercise 9.2
(1) Simplify the following:
(i) log a3 – log a2
Solution:
loga3 – loga2
= 3 log a – 2 log a
= log a
(ii) log a3 ÷ log a2
Solution:
log a3 ÷ log a2
= 3log a/2log a
= 3/2
(iii) log 4/log 2
Solution:
log 4/log 2
= log22/log2
= 2log 2/log 2
= 2
(iv) log 8 log 9/log 27
Solution:
log 8 log 9/log 27
= log 23 log 32/log 33
= 3log 2 . 2log 3/3log 3
= 2log 2
= log 22
= log 4
(v) log 27/log √3
Solution:
log 27/log √3
= log 33/log 31/2
= 3log 3/(1/2 log 3) = 6
(vi) log 9 – log 3/log27
Solution:
log9 – log3/log27
= log 32 – log 3/log 33
= 2 log 3 – log 3/3log 3
= log 3/3log 3
= 1/3
(2) Evaluate the following:
(i) log (10 ÷ ∛10)
Solution:
log (10 ÷ ∛10)
= log 10 – log ∛10
= log 10 – log 101/3
= log 10 – 1/3 log 10
= log 10 (1 – 1/3)
= log 10 (2/3)
= 2/3 × 1 [∵ log 10 = 1]
= 2/3
(ii) 2 + 1/2 log (10-3)
Solution:
2 + 1/2 log (10-3)
= 2 + ½ × -3 log 10
= 2 – 3/2 log 10
= 4-3/2 × 1 [∵ log 10 = 1]
(iii) 2log 5 + log 8 – 1/2 log 4
Solution:
2log 5 + log 8 – 1/2 log 4
= 2 log 5 + log 23 – log 22×1/2
= 2 log 5 + 3log2 – log2
= 2 log 5 + 2 log 2
= log 52 + log22
= log 25 + log 4
= log 25 × 4
= log 100
= 2 [∵ log10 100 = 2]
(iv) 2 log 103 + 3 log 10-2 – 1/3 log 5-3 + 1/2 log 4
Solution:
2 log 103 + 3 log 10-2 – 1/ log 5-3 + 1/2 log 4
= 6 log 10 + 3 × (-2) log 10 – log 5-3×1/3 + log 23×1/2
= 6 log 10 – 6 log 10 – log 5-1 + log 2
= – (-1) log 5 + log 2
= log 5 + log 2
= log 5 × 2
= log 10
= 1
(v) 2 log 2 + log 5 – 1/2 log 36 – log 1/30
Solution:
2 log 2 + log 5 – 1/2 log 36 – log 1/30
= 2 log 2 + log 5 – log 62×1/2 – log 1 – log 30
= 2 log 2 + log 5 – log 6 – 0 – log 3×10
= 2 log 2 + log 5 – log 2×3 – log 3 + log 10
= 2 log 2 + log 5 – log 2 + log 3 – log 3 + 1
= log 2 + log 5 + 1 [∵ log 1 = 0, log 10 = 1]
= log 2 × 5 + 1
= log 10 + 1
= 1 + 1
= 2
(vi) 2 log 5 + log 3 + 3log 2 – 1/2 log 36 – 2 log 10
Solution:
2 log 5 + log 3 + 3log 2 – 1/2 log 36 – 2 log 10
= 2 log 5 + log 3 + 3 log 2 – log 62×1/2 – 2×1
= 2 log 5 + log 3 + 3 log 2 – log 6 – 2
= 2 log 5 + log 3 + 3log2 – log 2×3 – 2
= 2 log 5 + log 3 + 3 log 2 – log 2 – log 3 – 2
= 2 log 5 + log 3 + 3 log 2 – log 2 – log 3 – 2
= 2 log 5+ 2 log 2 – 2
= 2 (log 10) – 2
= 2 – 2 = 0
(vii) log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80
Solution:
log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80
= log 2 + log (16/15)16 + log (25/24)12 + log (81/80)7
= log {2 × (16/15)16 × (25/24)12 × (81/80)7}
= log (21 × 24×16/316 × 516 × 52×12/23×12×312 × 34×7/24×7×57)3 — (i)
= log (21+64-36-28 × 328-16-12 × 524-16-7)
= log (21 × 3 × 51)
= log 2 × 5 × 1
= log 10
= 1
(viii) 2log 5 + log 8 – 1/2 log 4
Solution:
2log 5 + log 8 – 1/2 log 4
= log 52 + log 8 – log 22×1/2
= log 25 + log 8 – log 2
= log (25×8/2)
= log 100
= 2
(3) Express each of the following as a single logarithm:
(i) 2 log 3 – 1/2 log 16 + log 12
Solution:
2 log 3 – 1/2 log 16 + log 12
= log 32 – log 24×1/2 + log 12
= log 9 – log 4 + log 12
= log (9×12/4)
= log 27
(ii) 2log10 5 – log10 2 + 3log10 4 + 1
Solution:
2log 5 – log 2 + 3 log 4 + 1
= log 52 – log 2 + log 43 + log 10
= log 25 – log 2 + log 64 + log 10
= log (25×64×10/2)
= log 8000
(iii) 1/2 log 36 + 2 log 8 – log 1.5
Solution:
1/2 log 36 + 2 log 8 – log 1.5
= log 62×1/2 + log 82 – log 15/10
= log (6 × 64 × 18/15)
= log 256
(iv) 1/2 log 25 – 2 log 3 + 1
Solution:
1/2 log 25 – 2 log 3 + 1
= log25 2 x 1/2 – log32 + log10
= log (5×10/9)
= log 50/9
(v) 1/2 log 9 + 2 log 3 – log 6 + log 2 – 2
Solution:
1/2 log 9 + 2 log 3 – log 6 + log 2 – 2
= log 32×1/2 + log 32 – log 6 + log 2 – log 100
= log (3×9×2/6×100)
= log 9/100
(4) Prove the following:
(i) log10 4 ÷ log10 2 = log3 9
Solution:
(i) To Prove,
Log10 4 ÷ log10 2 = log3 4
L.H.S
Log10 4/log102
= 2log102/log102
= 2
RHS
Log39
= log3 32
= 2log3 3
= 2 × 1
= 2
∴ LHS = RHS
(ii) log10 25 + log10 4 = log5 25
Solution:
(ii) To Prove, log10 25 + log10 4 = log5 25
LHS
Log10 25 + log10 4
= log10 25 × 4
= log10 100
= 2
RHS
Log5 25
= log5 5
= 2 log5 5
= 2 × 1
= 2
∴ LHS = RHS
(5) If x = (100)a, y = (10000)b and z = (10)c, express log 10√y/x2z3 in terms of a, b, c.
Solution:
Given, x = (100)a, y = (10000)b z = (10)c
∴ log x = log 100a
Or, log x = a log 100
Log y = log 10000b
Or, log y = b log 10000
Log z = log 10c
Or, log z = c log 10
Now, log 10√y/x2z3
= log 10√y – log x2z3
= log 10 + log y1/2 – (log x2 + log z3)
= 1 + 1/2 log y – 2log x – 3 log z
Putting the values of log x, log y log z we get,
= 1 + 1/ b log 10000 – 2a log 100 – 3c log 10
= 1 + 1/2 b × 4 – 2a × 2 – 3c × 1
= 1 + 2b – 4a – 3c
(6) If a = log10 x, find the following in terms of a:
(i) x
(ii) log10 ∛x2
Solution:
Given,
A = log10 x
(i) a = log10 x
∴ x = 10a
(ii) log10 5√x2
= log10 x2×1/5
= log10 × 2/5
= 2/5 log10 x
= 2/5 × 9
= 24/5
(7) If a = log 2/3, b = log 3/5 and c = 2 log √5/2, find the value of
(i) a + b + c
(ii) 5a+b+c
Solution:
Given, a log 2/3, b = log 3/5, c = 2log √5/2
(i) a + b + c
= log 2/3 + log 3/5 + 2 log √5/2
= log 2/3 + log 3/5 + log (5/2)1/2×2
= log (2/3 × 3/5 × 5/2)
= log 1
= 0
(ii) 5a+b+c
= 50 [∵ a + b + c = 0 provide only found]
= 1
(8) If x = log 3/5, y log 5/4 and z = 2 log √3/2, find the values of
(i) x + y – z
(ii) 3x+y-z
Solution:
Given, x = log 3/5, y = log 5/4, z = 2 log √3/2
(i) x + y – z
(9) If x = log10 12, y = log4 2 × log10 9 and z = log10 0-4, find the values of
(i) x – y – z
(ii) 7x-y-z
Solution:
Given, x = log10 12, y = log4 2 × log10 9
Z = log100.4
Or, z = log10 4/10
Or, y = log4 41/2 × log10 9
= 1/2 × log4 4 × log10 9
= 1/2 × 1 × log10 9
= log10 3
y = log10 3
(i) x – y – z
= log10 12 – log10 3 – log10 4/10
= log10 (12×10/3×4)
= log10 10
= 1
(ii) 7x-y-z
= 71
= 7
(10) If log V + log 3 = log π + log 4 + 3 log r, find V in terms of other quantities.
Solution:
Given, log V + log 3 = log π + log 4 + 3 log r
Or, log V × 3 = log (π × 4 × r3)
Or, log 3V = log 4πr3
Or, 3V = 4 πr3
Or, V = 4/3 πr3
(11) Given 3 (log5 – log3) – (log5 – 2log 6) = 2 – log n, find n.
Solution:
Given,
3 (log 5 – log 3) – (log 5 – 2 log 6) = 2 – log n.
Or, 3 (log 5/3) – (log 5 – log 62) = log 100 – log n
Or, log (5/3)3 – log 5/36 = log 100 – log n
Or, log n = log 100 – log 125/27 + log 5/36
Or, log n = log (100×5×27/125×36)
Or, log n = log 3
∴ n = 3
(12) Given that log10 y + 2log10 x = 2, express y in terms of x.
Solution:
Given, log10 y + 2log10 x = 2
Or, log10 y = log10 100 – 2log10 x
Or, log10 y = log10 100 – log10 x2
Or, log10 y = log10 (100/x2)
∴ y = 100/x2
(13) Express log10 2 + 1 in the form log10 x.
Solution:
Log10 2 + 1
= log10 2 + log10 10
= log10 2 × 10
= log10 20
(14) If a2 = log10 x, b3 = log10 y and a2/2 – b3/3 = log10 z, express z in terms of x and y
Solution:
Given, a2 = log10 x, b3 = log, 10y
a2/2 – b3/3 = log 10 z
Or, log10 x/2 – log10 y/3 = log 10 z
Or, 3 log 10 x – 2 log 10 y/6 = log10 z
(15) Given that log m = x + y and log n = x – y, express the value of log m2n in terms of x and y.
Solution:
Given, log m = x + y, log n = x – y
∴ log m2x
= log m2 + log n
= 2 log m + log n
= 2 (x + y) + (x – y)
= 2x+ xy + x – y
= 3x + y
(16) Given that log x = m + n and log y = m – n, express the value of log (10x/y2) in terms of m and n.
Solution:
Given, log x = m + n, log y = m – n
∴ log 10x/y2
= log 10x – log y2
= log 10 + log x – 2 log y
= 1 + m + n – 2 (m – n)
= 1 + m + n – 2m + 2n
= 1 + 3n – m
(17) If log x/2 = log y/3, find the value of y4/x6
Solution:
Given, log x/2 = log y/3
Or, 3 log x = 2 log y
Or, log x3 = log y2
Or, x3 = y3
Or, x3×2 = y2×2
Or, x6 = y4 — (i)
∴ y4/x6
= y4/y4 [From (i)]
= 1
(18) Solve for x:
(i) log x + log 5 = 2 log 3
Solution:
log x + log 5 = 2 log 3
Or, log x = log 32 – log 5
Or, log x = log 9/5
∴ x = 9/5
(ii) log3 x – log3 2 = 1
Solution:
log3 x – log3 2 = 1
Or, log3 x = 1 + log3 2
Or, log3 x = log3 3 + log3 2
Or, log3 x = log3 3×2
Or, log3 x = log3 6
∴ x = 6
(iii) x = log 125/log 25
Solution:
x = log 125/log 25
x = log 53/log 52
Or, x = 3 log5/2log 5
Or, x = 3/2
(iv) log 8/log 2 × log 3/log √3 = 2 log x
Solution:
log 8/log 2 × log 3/log √3 = 2 log x
Or, log 23/log 2 × log 3/log 31/2 = 2 log x
Or, 3log 2/log 2 × log 3/log 3 = 2log x
Or, 2 log x = 3 × 2
Or, log x = 3
Or, log x = log 1000
∴ x = 1000
(19) Given 2 log10 x + 1 = log10 250, find
(i) x
(ii) log10 2x
Solution:
(20) If logx/log5 = log y2/log 2 = log 9/(log 1/3), find x ad y.
Solution:
Given, logx/log5 = log y2/log 2 = log 9/(log 1/3),
∴ log x/log 5 = log 9/(log 1/3)
Or, log x/log 5 = log 32/log 1 – log 3
Or, log x/log 5 = 2 log 3/0 – log 3
Or, log x/log 5 = 2log 3/log 3
Or, log x = – 2 log 5
Or, log x = log 5-2
∴ x = 5-2
Or, x = 1/25
Also, log y2/log 2 = log 9/log 1/3
Or, log y2/log 2 = log 33/log 1 – log 3
Or, log y2/log 2 = 2 log 3/0 – log 3
Or, log y2/log 2 = 2 log 3/log 3
Or, log y2 = – 2 log 2
Or, log y2 = log 2-2
∴ y = 2-2
Or, y = 1/2
(21) Prove the following:
(i) 3log4 = 4log3
(ii) 27log2 = 8log3
Solution:
(i) 3log4 = 4log3
Putting log on sides are get,
Log3log4 = log4log3
Or, log 4 log 3 = log 3 by 4
∴ LHS = RHS
(ii) 27log2 = 8log3
Putting log on both sides we get,
Log 27log2 = log 8log3
Or, log 2 log 33 = log 3.log 23
Or, 3 log 3 . log 2 = 3 log 3. Log 2
∴ LHS = RHS
(22) Solve the following equations:
(i) log (2x + 3) = log 7
Solution:
log (2x + 3) = log 7
∴ 2x + 3 = 7
Or, x = 4/2 = 2
(ii) log (x + 1) + log (x – 1) = log 24
Solution:
log (x + 1) + log (x – 1) = log 24
Or, log (x + 1) (x – 1) = 24
∴ (x + 1) (x – 1) = 24
Or, x2 – 1 = 24
Or, x2 = 25
Or, x = √25 = 5
(iii) log (10x + 5) – log (x – 4) = 2
Solution:
Log (10x + 5) – log (x – 4) = 2
Or, log (10x + 5)/x-4 = log 100
∴ 10x + 5/x-4 = 100
Or 10x + 5 = 100x – 400
Or, 100x – 10x = 400 + 5
Or, 90x = 405
Or, x = 405/90 = 45/10 = 4.5
(iv) log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
Solution:
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
Or, log10 5 (5x + 1) = log10 (x + 5) + log10 10
Or, log10 (25x + 5) = log10 (10x + 50)
∴ 25x + 5 = 10x + 50
Or, 15x = 45
Or, x = 3
(v) log (4y – 3) = log (2y + 1) – log 3
Solution:
log (4y – 3) = log (2y + 1) – log 3
Or, log (4y – 3) = log (2y + 1/3)
∴ 4y – 3 = 2y + 1/3
Or, 12y – 9 = 2y + 1
Or, 10y = 10
Or, y = 1
(vi) log10 (x + 2) + log10 (x – 2) = log10 3 + 3 log10 4
Solution:
log10 (x + 2) + log10 (x – 2) = log10 3 + 3 log10 4
=> log10 (x + 2) (x – 2) = log10 3 + log10 43
Or, log10 (x2– 22) = log10 3 × 64
∴ x2 – 4 = 3 × 64
Or, x2 = 192 + 4
Or, x = √196
Or, x = 14
(vii) log (3x + 2) + log (3x – 2) = 5 log 2
Solution:
Log (3x + 2) + log (3x – 2) = 5log2
Or, log (3x + 2) (3x – 2) = log25
Or, log (3x)2 – 22 = log 32
∴ 9x2 – 4 = 32
Or, 9x2 = 36
Or, x2 = 4
Or, x = √4
Or, x = 2
(23) Solve for x: log3 (x + 1) – 1 = 3 + log3 (x – 1)
Solution:
Log3 (x + 1) – 1 = 3 + log3 (x – 1)
Or, log3 (x + 1) – log3 3 = log3 27 + log3 (x + 1)
Or, log3 (x+1)/3 = log3 27 (x – 1)
[∵ 3 = 1 + 1 + 1
= log3 3 + log3 3 + log3 3
= log3 3×3×3
= log3 27]
∴ x+1/3 = 27 (x – 1)
Or, x + 1 = 81x – 81
Or, 80x = 82
Or, x = 82/80
Or, x = 41/40
(24) Solve for x: 5log x + 3logx = 3log x + 1 – 5log x – 1
Solution:
5log x + 3logx = 3log x + 1 – 5log x – 1
Or, 5logx + 5logx-1 = 3logx+1 – 3logx
Or, 5logx + 5logx × 5-1 = 3logx × 31 – 3logx
Or, 5logx (1 + 1/5) = 3logx (3 – 1)
Or, 5logx/3logx = 2/(5+1/5)
Or, 5logx/3logx = 10/6
Or, 5logx/3logx = (5/3)
Putting both sides we get,
Log x = 1
Or, log x = log 10
∴ x = 10
(25) If log x-y/2 = 1/2 (log x + log y), Prove that x2 + y2 = 6xy
Solution:
To prove,
X2 + y2 = 6xy
Now, log x-y/2 = 1/2 (log x + log y)
Or, log x-y/2 = 1/2 log xy
Or, log x-y/2 = log (xy)1/2
∴ x-y/2 = (xy)1/2
Or, (x-y/2)2 = xy [Squaring both sides]
Or, x2 – 2xy + y2/4 = xy
Or, x2 + y2 = 4xy + 2xy
Or, x2 + y2 = 6xy (Hence Proved)
(26) If x2 + y2 =23xy, prove that log x+y/5 = 1/2 (log x + log y)
Solution:
Given, x2 + y2 = 23xy
To prove, log x+y/5 = 1/2 (log x – log y)
Now, x2 + y2 = 23xy
Or, x2 + y2 = 25xy – 2xy
Or, x2 + 2xy + y2 = 25xy
Or, (x + y)2 = 25xy
Or, x + y = √25xy
Or, x + y = 5 (xy)1/2
Or, x + y/5 = (xy)1/2
Putting log on both sides we get,
Or, log x + y/5 = log (xy)1/2
Or, log x+y/5 = 1/5 log xy
Or, log x+y/5 = 1/2 (log x + log y) (Hence Proved)
(27) If p = log10 20 and q = log10 25, find the value of x if
2 log10 (x + 1) = 2p – q
Solution:
Given, p = log10 20 & q = log10 25
Now, 2 log10 (x + 1) = 2p – q
Or, log10 (x + 1)2 = 23 log10 20 – log10 25
Or, log10 (x + 1)2 = log10 202 – log10 25
Or, log10 (x + 1)2 = log 202/25
∴ (x + 1)2 = 202/25
Or, (x + 1)2 = (20/5)2
Or, (x + 1) = 20/5
Or, x = 20/5 – 1
Or, x = 15/5
Or, x = 3
(28) Show that:
(i) 1/log2 42 + 1/log3 42 + 1/log7 42 = 1
Solution:
(i) To Prove,
1/log2 42 + 1/log3 42 + 1/log7 42 = 1
LHS
1/log242 + 1/log342 + 1/log7 42
= log42 2 + log42 3 + log42 7 [reciprocal low]
= log42 (2×3×7)
= log42 42
= 1
= RHS/Proved
(ii) 1/log8 36 + 1/log9 36 + 1/log18 36 = 2
LHS
1/log8 36 + 1/log9 36 + 1/log18 36
= log36 8 + log36 9 + log36 18
= log36 (8×9×18)
= log36 1296
= log36 36×36
= log36 36 + log36 36
= 1 + 1 = 2 = RHS/Proved
(29) Prove the following identities:
(i) 1/loga abc + 1/logb abc + 1/logc abc = 1
(ii) logb a . logc b . logd c = logd a.
Solution:
(i) To Prove,
1/loga abc + 1/logb abc + 1/logc abc = 1
LHS
1/loga abc + 1/logb abc + 1/logc abc = 1
= logabc a + logabc b + logabc c [reciprocal]
= logabc a×b×c
= lagabc abc
= 1 (RHS Prove)
(ii) To prove, logb a . logc b . logd c = logd a
LHS
Logb a . logc b . logd c
= (logb a . logc b) . logd c
= logc a . logd c [reciprocal multiplication low]
= logd a [reciprocal multiplication low]
= RHS Proved
(30) Given that loga x = 1/α, logb x = 1/β, logc x = 1/γ, find logabc x.
Solution:
loga x = 1/α, logb x = 1/β, logc x = 1/γ
Putting log on both sides we get,
∴ 1/ α = loga x = logx/log a
Or, log a = α log x
Similarly, 1/β = logb x = logx/log b
Or, log b = β log x
1/γ = logc x = logx/log c
Or, log c = γ log x
Now, logabc x = log x/log abc
= log x/loga + logb + logc
= log x/α log x + β log b + γ log c
Or, logabc x = log x/log x (α+β+γ)
Or, logabc x = 1/ α+β+γ
(31) Solve for x:
(i) log3 x + log9 x + log81 x = 7/4
Solution:
log3 x + log9 x + log81 x = 7/4
Or, log x/log 3 + log x/log 9 + log x/log 81 = 7/4
Or, log x/log 3 + log x/2 log 3 + log x/4 log 3 = 7/4
Or, 4log x + 2log x + log x/4log 3 = 7/4
Or, 7log = 7log 3
∴ x = 3
(ii) log2 x + log8 x + log32 x = 23/15
Solution:
Log2 x + log8 x + log32 x = 23/15
Or, log x/log 2 + log x/log 8 + log x/log 32 = 23/15
Or, log x/log 2 + log x/3log 2 + log x/5log 2 = 23/15
Or, 15log x + 5 log x + 3log x/15 log 2 = 23/15
Or, 23log x = 23 log 2
Or, log x = log 2
∴ x = 2
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