Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8D Solutions
EXERCISE – 8D
(1.) If 3/2 log a + 2/3 log b – 1 = 0, find the value of a9 b4.
Solution :
3/2 log a + 2/3 log b – 1 = 0
3/2 log a + 2/3 log b = 1
9 log a + 4 log b / 6 = 1
9 log a + 4 log b = 6
log a9 + log b4 = 6
log10 (a9 b4) = 6
106 = a9 b4
(2) If x = 1 + log2 – log5, y = 2 log3 and z = log a – log5; find the value of a
Solution :
Given that,
x = 1 + log 2 – log 5
y = 2 log 3
and Z = log a – log 5
Now, we have to find the value of a.
x + y = 2z
1 + log 2 – log 5 + 2 log 3 = 2 (log a – log 5)
log10 + log2 – log5 + log32 = 2 log a – 2 log5
log10 + log2 – log5 + log9 = loga2 – log52
log (10×2×9/5) = log (a2/52)
log (36) = log (a2/25)
36 = a2/25
36 × 25 = a2
Taking square root on both sides,
√36 × 25 = √a2
6 × 5 = a
30 = a
∴ a = 30
(3) If x = log 0.6, y = log 1.25 and z = log 3 – 2 log 2, find the values of :
(i) x + y – z
Solution :
x + y – z
Given that, x = log 0.6
y = log 1.25
and z = log 3 – 2 log 2
we have to putting the value of x, y and z in given equation.
x + y – z
=> log 0.6 + log 1.25 – (log 3 – 2 log 2)
=> log 0.6 + log 1.25 – log 3 + log 22
=> log (0.6 × 1.25 × 4/3)
=> log (0.6 × 1.25 × 4 / 3 × 1000)
=> log (8/8)
=> log 1
= 0
(ii) 5x+y-z
Solution :
5x+y-z
We have already find out the value of x + y – z = 0
So, 50
= 1 {∵ a0 = 1}
(4) If a2 = log x, b3 = log y and 3a2 – 2b3 = 6 log z, express y in terms of x and z.
Solution :
Given that,
a2 = log x
b3 = log y
and 3a2 – 2b3 = 6 log z
first we have to putting the value of a2 and b3 in given equation.
3 log x – 2 log y = 6 log z
log x3 – log y2 = log z6
log x3/y2 = log z6
x3/y2 = z6
x3/z6 = y2
Taking square root on both sides,
√x3/z6 = √y2
y = √x3/z6
(5) If log (a – b)/2 = 1/2 (log a + log b), show that a2 + b2 = 6ab
Solution :
Given that,
log (a – b)/2 = 1/2 (log a + log b)
log (a – b)/2 = log (a)1/2 + log (b)1/2
log (a – b)/2 = log (√a) + log (√b)
log (a – b)/2 = log (√a × √b)
a – b/2 = (√a × √b)
a – b = 2 √ab
Squaring on both sides,
(a – b)2 = (2 √ab)2
a2 + b2 – 2ab = 4ab
a2 + b2 = 4ab + 2ab
a2 + b2 = 6ab
Hence proved.
(6) If a2 + b2 = 23ab, show that;
log (a + b/5) = 1/2 (log a + log b)
Solution :
Given that,
a2 + b2 = 23ab
adding 2ab on both sides,
a2 + b2 + 2ab = 23ab + 2ab
(a + b)2 = 25ab
Taking log on both sides,
log (a + b)2 = log 25ab
2 log (a + b) = log 25 + log a + log b
2 log (a + b) = log 52 + log a + log b
2 log (a + b) = 2 log 5 + log a + log b
2 log (a + b) – 2 log 5 = log a + log b
2 [log (a + b) – log 5] = log a + log b
log (a + b/5) = 1/2 (log a + log b)
Hence proved.
(7) If m = log 20 and n = log 25, find the value of x, so that; 2 log (x – 4) = 2m – n
Solution :
Given that,
m = log 20
and n = log 25
∴2 log (x – 4) = 2m – n
putting the value of m and n in given equation.
2 log (x – 4) = 2 log 20 – log 25
log (x – 4)2 = log (20)2 – log 25
log (x – 4)2 = log [20 × 20/25]
log (x – 4)2 = log (16)
(x – 4)2 = 16
Taking square root on both sides,
√(x – 4)2 = √16
x – 4 = 4
x = 4 + 4
x = 8
(8) Solve for x and y; if x > 0 and y > 0 ; log xy = log x/y + 2 log 2 = 2.
Solution :
log xy = log x/y + 2 log 2 = 2
log xy = log x/y + log 22
log xy = log x/y + log 4
log xy = log 4x/y
xy = 4x/y
y = 4/y
y2 = 4
Taking square root on both sides,
√y2 = √4
y = 2 ……(i)
log xy = 2
102 = xy
100 = xy
100 = x (2) {from (i)}
100/2 = x
50 = x
(9) Find x, if ;
(i) logx 625 = – 4
Logx 625 = – 4
x-4 = 625
1/x4 = 625
x4 = 1/625
Taking 4th root on both sides,
4√x4 = 4√1/625
x = (1/54)1/4
x = 1/5
(ii) logx (5x – 6) = 2
Solution :
logx (5x – 6) = 2
x2 = 5x – 6
x2 – 5x + 6 = 0
x2 – 3x – 2x + 6 = 0
x (x – 3) – 2 (x – 3) = 0
(x – 3) (x – 2) = 0
x – 3 = 0 or x – 2 = 0
x = 3 or x = 2
(10) If p = log 20 and q = log 25, find the value of x, if 2 log (x + 1) = 2p – q
Solution :
Given that,
p = log 20 and
q = log 25
2 log (x + 1) = 2p – q
2 log (x + 1) = 2 log 20 – log 25 (∵ Given)
2 log (x + 1) = 2 log 20 – log 52
2 log (x + 1) = 2 log 20 – 2 log 5
2 log (x + 1) = 2 [log20 – log5]
2/2 log (x + 1) = log 20 – log 5
log (x + 1) = log (20/5)
log (x + 1) = log (4)
x + 1 = 4
x = 4 – 1
x = 3
(11) If log2 (x + y) = log3 (x – y) = log 25/log 0.2, find the value of x and y.
Solution :
Given that,
log2 (x + y) = log3 (x – y) = log25/log 0.2
log2 (x + y) = log 25/log 0.2
log2 (x + y) = log 52/log (2/10)
log2 (x + y) = 2 log 5/log 1/5
log2 (x + y) = 2log 5/log 5-1
log2 (x + y) = 2log 5/ – log 5
log2 (x + y) – 2
2-2 = (x + y)
1/22 = x + y
1/4 = x + y ….. (i)
Now, log3 (x – y) = log 25/log 0.2
log3 (x – y) = log 52/log(2/10)
log3 (x – y) = 2 log 5/log (1/5)
log3 (x – y) = 2 log 5/log (5-1)
log3 (x – y)= 2 log5/- log5
log3 (x – y) = – 2
3-2 = x – y
1/32 = x – y
1/9 = x – y…… (ii)
Adding equation (i) and (ii),
we get,
x + y = 1/4
x – y = 1/9
2x = 1/4 + 1/9
2x = 9 + 4/36
2x = 13/36
x = 13/2 × 36
x = 13/72
Put x = 13/72 in equation (i),
we get,
13/72 + y = 1/4
y = 1/4 – 13/72
y = 18 – 13/72
y = 5/72
∴ The value of x = 13/72 and y = 5/72
(12) Given log x/log y = 3/2 and log xy = 5;
Find the values of x and y
Solution :
Given that,
log x/log y = 3/2
2 log x = 3 log y
log x2 = log y3
x2 = y3 …. (i)
Now, log xy = 5
105 = xy
105/y = x
putting x = 105/y in equation (i),
we get,
(105/y)2 = y3
1010/y2 = y3
1010 = y3 y2
1010 = y5
1010 = y5
(1010)1/5 = y
102 = y
y = 100
put y = 100 in equation (ii),
we get,
x = 105/100
x = 100000/100
x = 1000
(13) Given log10 x = 2a and log10 y = b/2
(i) Write 10a in terms of x.
Solution :
log10 x = 2a
102 a = x
(10a)2 = x
(10a)2 = x
Taking square root on both sides
10a = √x
(ii) 102b+1 in terms of y
Solution :
log10 y = b/2
10b/2 = y
Taking 4th power on both sides,
104×b/2 = y4
102b = y4
102b × 101 = y4 × 101
102b+1 = 10y4
(iii) If log10 p = 3a – 2b, express p in terms of x and y
Solution :
Given that,
log10 p = 3a – 2b
log10 p = 3 [log10 x/2] – 2[2log10 y] (Given)
log10 p = log10 x3/2 – log10 y4
log10 p = log10 x3/2/y4
p = x3/2/y4
(14) Solve : log5 (x + 1) – 1 = 1 + log5 (x – 1)
Solution :
Given equation,
log5 (x + 1) – 1 = 1 + log5 (x – 1)
log5 (x + 1) – log5 (x – 1) = 1 + 1
log5 [(x + 1)/(x – 1)] = 2
=> 52 = x + 1/x – 1
=> 25 = x + 1/x – 1
=> 25 (x – 1) = x + 1
=> 25x – 25 = x + 1
=> 25x – x = 1 + 25
=> 24x = 26
x = 26/24
x = 13/12
(15) Solve for x, if ;
logx 49 – logx 7 + logx 1/343 + 2 = 0
Solution :
logx 49 – logx + 7 + logx 1/343 + 2 = 0
logx 72 – logx7 + logx (343)-1 = -2
logx 49 – logx 7 – logx (343) = -2
logx [49/7×343] = – 2
logx [1/49] = – 2
x-2 = 1/49
1/x2 = 1/49
x2 = 49
Taking square root on both sides,
√x2 = √49
x = 7
(16) If a2 = log x, b3 = log y and a2/2 –b3/3 log c, find c in terms of x and y.
Solution :
Given that,
a2 = log x
b3 = log y
and
a2/2 – b3/3 = log c
putting the values of a2 and b3 in given equation,
log x/2 – log y/3 = log c
3 log x – 2 log y/6 = log c
3 log x – 2 log y = 6 log c
log x3 – log y2 = log c6
log x3/y2 = log c6
x3/y2 = c6
6√x3/y2 = c
(17) Given : x = log10 12, y = log4 2 × log10 9 and z = log10 0.4 find
(i) x – y – z
Solution :
Given that,
x = log10 12
y = log4 2 × log10 9
and z = log10 0.4
putting the values of x, y, and z in given equations
x – y – z
= log10 12 – [log4 2 × log10 9] – log10 0.4
= log10 (4×3) – [log2/log4 × log10 32] – log10 (4/10)
= log10 4 + log10 3 – (log2/log22 × 2 log10 3) – (log10 4 – log10 10)
= log10 4 + log10 3 – (log2/2 log2 × 2 log10 3) – log10 4 + log10 10
= log10 4 + log10 3 – log10 3 – log10 4 + log10 10
= log10 10
= 1
(ii) 13x-y-z
Solution : 13x-y-z
We have already find out the value of x – y – z is 1
so, 131
= 13
(18) Solve for x, logx 15√5 = 2 – logx 3√5
Solution :
logx 15√5 = 2 – logx 3√5
logx 15√5 + logx 3√5 = 2
logx (15√5 × 3√5) = 2
logx (45 × 5) =2
x2 = 45 × 5
x2 = 225
Taking square root on both sides,
√x2 = √225
x = 15
(19) Evaluate :
(i) logb a × logc b × loga c
Solution :
logb a × logc b × loga c
= loga/logb × logb/logc × logc/loga
= 1
(ii) log3 8 ÷ log9 16
Solution :
log3 8 ÷ log9 16
= log3 8/log9 16
= log 8/log 3/log 16/log 9
= log 23/log3 × log32/log24
= 3 log2/log3 × 2 log3/4 log2
= 3/2
(iii) log5 8/(log25 16 × log100 10)
Solution :
log5 8/(log25 16 × log100 10)
= log8/log5 / log16/log25 × log10/log100
= log23/log5 / log24/log52 × log10/log102
= 3 log 2/log 5 / 4 log2/2 log5 × log10/2 log10
= 3log2/log5 / 4 log2/4 log5
= 3 log2/log5 × log5/log2
= 3
(20) Show that :
loga m ÷ logab m = 1 + loga b
Proof :
L.H.S. = loga m ÷ logab m
= loga m/logab m
= logm/loga/logm/logab
= log m/log a × log ab/log m
= log ab/log a
= loga ab
= loga a + loga b
= 1 + loga b
L.H.S. = R.H.S.
Hence proved.
(21) If log√27 x = 2 2/3, find x
Solution :
log√27 x = 2 2/3
log√27 x = 8/3
log x/log √27 = 8/3
log x/log (27)1/2 = 8/3
log x/log (33)1/2 = 8/3
log x/log (3)3/2 = 8/3
log x/ 3/2 log 3 = 8/3
log x/log 3 = 8/3 × 3/2
log x/log 3 = 4
log x = 4 log 3
log x = log 34
x = 34
x = 81
(22) Evaluate :
1/(loga bc + 1) + 1/(logb (a+ 1)) + 1/(logc ab + 1)
Solution :
1/(loga bc + 1) + 1/(logb (a+ 1)) + 1/(logc ab + 1)
= 1/loga bc + loga a + 1/logb ca + logb b + 1/logc ab + logc c
= 1/loga abc + 1/logb abc + 1/logc abc
= 1/ log abc/loga + 1/ log abc/logb + 1/ log abc/c
= log a/log abc + log b/log abc + log c/log abc
= log a + log b + log c/log abc
= log (abc)/log (abc)
= 1
Here is your solution of Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8D
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