Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Logarithmic equation Page number 92 (Q. No. 125 to 261) 125.
⇒ ⇒ ⇒
x – 1 ≠ 1 and x≠2
logx – 1 3 = 2 (x – 1)2 = 3
x–1>0 x>1
x2 + 1 – 2x = 3 x2 – 2x – 2 = 0 D = b2 – 4ac = 12 Roots =
−b+ D 2±2 3 2(1 ± 3 ) = = =1± 2a 2 2
3 ∴ Final solution x ∈ {
Since 1+ 3 satisfies the equation
}
126.
log4(2 log3(1 + log2(1 + 3 log3x))) =
1 2
⇒
1 1 log2(2 log3 (1 + log2 (1 + 3 log3x))) = 2 2
⇒ ⇒ ⇒ ⇒
2 log3 (1 + log2(1 + 3 log3x))) = 21 = 2
⇒ ⇒ ⇒
(1 + log2 (1 + 3 log34) = 31
127.
log3 (1 + log3 (2x – 7) = 1
log2 (1 + 3 log3x) = 3 – 1 3 log3x = 3 x=
31
1 + 3 log3x = 22 log3x = 1 ∴ Final solution x ∈ {3}
=3
⇒
2x = 9 + 7 = 16
128.
log3 (3x – 8) = 2 – x
⇒
⇒ ⇒
32 – x = 3x – 8
⇒
9 = t2 – 8t
(2x – 7) = 32
∴ Final solution x
x=4
Let 3x = t 9 =t–8 t
⇒
1 + log3 (2x – 7) = 3
⇒
{4}
⇒
= 3x – 8
9 3x
= 3x – 8
∈12 log 3 2 (39 − 2 x ) 1+ 3 2t2x –38t − x– 9 = 0 3
⇒
D = b2 – 4ac = 64 + 36 = 100 t = 3x =
8 ± 10 18 2 = ,– = 9 ,–1 (negative value is not possible as ax >0 ) 2 2 2
3x = 9 ⇒ 3x = 32 ⇒ x = 2 129.
x≠3
=1
⇒ log2 (9 – 2x) = 3 – x Now,
2x
∴ Final solution x ∈ {2}
= 1, 8
⇒(t – 1) (t – 8) = 0
23
⇒ 9 – 2x = (2)3 – x
⇒ 9 – 2x =
⇒ 8 = 9t – ⇒t = 1, 8
⇒ – 9t + 8 = 0 ⇒ x = 0, 3 but x ≠ 3
t2
2
⇒
x
8 =9–t t
t2
∴ Final solution x ∈ {0} 130.
⇒ 131.
25 – 65 = 8x
⇒ (5 – x)2 = x2 – 2x + 65 ⇒ – 40 = 8x
log3 (log9x +
+ 9x) = 2x
∴ Final solution x {– 5} 1 1 1 ⇒ log3 (log9x + + 32x) = 2x ⇒ – = log3x 2 2 2
–1 = log3x
⇒x=
log5 – x (x2 – 2x + 65) = 2
1 3
Mathematics Online: It’s all about Mathematics
⇒ 25 – x2 – 10x = x2 – 2x + 65 ⇒x=–5
∴ Final solution x ∈ {
}
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 132.
⇒ ⇒
log3(x + 1) (x + 3) = 1
A. I. Prilepko Solution
31 = (x + 1) (x + 3)
3 = x2 + 4x + 3 0 = x2 + 4x ∴ x = –4, 0 since – 4 doesn’t satisfy equation 133.
134.
log7 (2x – 1) + log7 (2x – 7) = 1 Let 2x = t log7(t – 1) (t – 7) = 1 ⇒ 71 = t2 – 7t – t + 7
⇒ t = 0, 8
2x = 8 ⇒ x = 3 2x = 0 Impossible
∴ Final solution x
log 5 + log(x + 10) – 1 = log (2/x – 20) – log (2x – 1)
⇒ log (5 + x + 10) – log1010 = log
5( x + 10 ) 21x − 20 = log 10 2x − 1 ⇒ (2x – 1) (x + 10) = 42x – 40 ⇒ 2x2 – 19x – 42x – 10 + 40 = 0
∴ Final solution x ∈ {
x + 10 21x − 20 = 2 2x − 1 ⇒ 2x2 – x + 20x – 10 = 42x – 40 ⇒ 2x2 – 20x – 3x + 30 = 0 3 ⇒ x = , 10 2
, 10 }
1 1 1 log + log x + log 5 2 3 3
135.
1 – log 5 =
⇒
log10
⇒
1 log1023 = log10 x 2
⇒ log10 10 – log105 =
10 1 = [log102–1 + log10x + log1051/3] 5 3 3
x−
x35 ∈ 1⇒x8− =20 321 16 2 3 23 25x − 1
⇒x=
= log
1 2 1 x+ 8
⇒
1 x− 2
2
x
1 x2 x x x2 = x3 + + x2 – – – 8 8 4 2 2
⇒
x3 +
⇒
8x3 + x2 = 8x3 – 2x + 4x2 – 1
⇒
0 = 3x (x –1) +1(x – 1)
137.
3log3 log
⇒
log10 x – log10x + log102x – 3 = 0
⇒
⇒
log10x – 2 log10x + 2(log10x) 2 – 62 = 0
⇒
⇒
2(log10x)2 – 4 log10x + 3 log10x – 6 = 0 ⇒
⇒
log10x = 2
or
log10x = –
⇒
x = 102 = 100
or
x = 10–3/2
x
16 3
5
}
x+
1 2
1 [log102–1 + log10x + log1051/3] 3
1 log102–1 + log10x + log1051/3 3
1 1 1 1 1 log x − = log x + – log x + ⇒log x – log 2 2 8 2 2
x
log
⇒ log1023 =
5
∴ Final solution x ∈ {
log x –
{3}
⇒
⇒ (2x – 3) (x – 10) = 0
⇒
∴ Final solution x ∈ {0}
⇒ 0 = t2 – 8t
⇒ log
136.
⇒ 3 = x2 + x + 3x + 3 ⇒ 0 = x(x + 4)
=
x+ 1 2 1 x + 8
1 1 = log x + – log 2 2
2
⇒
1 +x 4 1 x+ 8
x2 +
x2 x−
x+
1 2
=
⇒ 8(8 x3 + x2) = 8(8x3 + 2x + 8x2 – 4x2 – 1 – 4x) ⇒ 0 = 3x2 – 2x – 1 1 ⇒x=– ,1 3 ⇒
– logx + log2x – 3 = 0
x−
⇒ 0 = 3x2 – 3x + x – 1 ∴ Final solution x ∈ { –
}
log x – log x + log2x – 3 = 0 1 log10x – log10x + (log10x)2 – 3 = 0 2 2(log10x)2 – log10x – 6 = 0
(2 log10x + 3) (log10x – 2) = 0 3 2
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A. I. Prilepko Solution
1 8
Mathematics Online: It’s all about Mathematics 2
( x − 2 ) + log( x − 2 ) 5 −12
A. I. Prilepko Solution
138.
( x − 2) log
⇒
log ( x − 2)log
⇒ ⇒ ⇒ ⇒
log (x – 2) (log2(x – 2)+ log (x – 2)5 – 12)) = log10102 log(x – 2) t [t2 + 5t – 12)] = 2t ⇒ t3 + 5t2 – 14t = 0 ⇒ t(t + 7)(t – 2) = 0 ⇒ t = 0, –7, 2 ⇒ x = 2 ,2 + 10–7 , 102
139.
9 log3 (1− 2 x ) = 5x2 – 5
⇒
3 2 log3 (1− 2 x ) = 5x2 – 5
⇒ ⇒ ⇒ ⇒ ⇒
3log3 (1−2 x ) = 5x2 – 5
2
(Let log (x-2) = t )
= 102 log (x – 2)
( x − 2 ) + log( x − 2 )( 5 −12 )
= log 102log (x – 2) t(t2 + 5t – 14) = 0 log (x-2) = 0, –7, 2
2
(1 – 2x)2 = 5x2 – 5 1 + 4x2 – 4x = 5x2 – 5 0 = 5x2 – 4x2 – 4x – 5 – 1 0 = x2 – 4x – 6 D = b2 – 4ac = 16 + 24 = 40 D = 2 10 α=
140.
⇒ ⇒ ⇒
⇒ ⇒ ⇒ ⇒
10
x1 + log x = 10x logxx1 + logx = log 10x (log10x) (1 + log10x) = log1010 + log10x ⇒ log10x + log102x = 1 + log10x log102x = 1 ∴
141.
−b+ D = 2 + 10 , 2 – 2a
log10 x = ±1 x = 10–1, 101
x2 logx = 10x2 log10x2.log x = log1010 + log10x2 2 log10x (log10x) = 1 + 2 log10x 2 log102x – 2 log10x – 1 = 0
Let
log10x = t
2t2 – 2t – 1 = 0 2t2 D = b2 – 4ac = 4 + 8 = 12 t=
D = 2 3
2(1 ± 3 ) 1± 3 2±2 3 = = 4 2 9
Now, log10x = t =
1+ 3 1− 3 , 2 2
1± 3 2
∴
x=
⇒
log10x =
⇒
log10x2 = 1± 3
1± 3 2
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A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics
⇒
x2 = 101±
142.
x
log x +5 3
log x .
3
,
101−
log x + 5 = log10105 + log10x 3 log x + 5 = 5 + log10x 3
⇒
log x .
⇒ ⇒ ⇒
log2x + 5 log10x = 15 + 3 log10x log102x + 2 log10x – 15 = 0 log10x (log10x + 5) – 3(log10x + 5) = 0 (log10x – 3) (log10x + 5) = 0 log10x = 3 x=
log10x = – 5
103
x = 10–5
143.
x log3 . x = 9
⇒ ⇒ ⇒
log3 x log 3.x = log39 log3x (log3x) = 21
⇒
log3x = ±
⇒
x= 3
144.
( x )log5 x −1 = 5
⇒
log5
⇒
1 log5x (log5x – 1) = 1 2
⇒
1 1 log52x – log5x = 1 2 2
⇒ ⇒ ⇒ ⇒ ⇒
= log52 x – log5x = 2
log5x = 2
and
⇒
x = 52
and
⇒
x=5
and
145.
xlogx +1 = 106 log x (log x + 1) = log10106
⇒ ⇒ ⇒ ⇒
3
= 105 + log x
⇒ ⇒
⇒
3
101+
x=
A. I. Prilepko Solution
log32x = 21
2
2
, 3−
2
x . log5 (x – 1) = log55
= log52 x – log5x – 2 = 0 log5x (log5x + 1) – 2(log5x + 1) = 0 (log5x – 2) (log5x + 1) = 0 log5x = –1 1 x= 5 1 x= 5
log102x + log10x – 6 = 0 (log10x – 2) (log10x + 3) = 0 log10x = 2
log10x = – 3
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics ⇒ x = 102 x = 10–3
146.
log x +7 x 4
⇒
log x. log
⇒
2 log10 x + 3 log10x – 4 = 0
⇒
2 log10 x + 4 log10x – log10x – 4 = 0
⇒ ⇒
(log10x + 4) (log10x – 1) = 0
147.
x log
⇒
x1/ 2 logx ( x −2 ) = 9
⇒
x logx ( x −2 ) = 9
⇒
( x − 2) 2 log x x = 9
= 10log x + 1 x+7 = 4 log10 x + 4 4
x = 104 and x
A. I. Prilepko Solution
( x −2)
x = 101
=9
2
x
⇒⇒ (x – 2)2 – 9 = 0 ⇒x2 + 4 – 4x – 9 = 0 ⇒ x2 – 4x – 5 = 0 ⇒ x(x – 5) + 1(x – 5) = 0 ⇒ x = –1, 5 but x > 0 ⇒ ∴ x= 5 is a solution log2 x + log x2 − 2
148.
log x 2
⇒
2 2 1 1 log x log x + log x − 2 = 2 log x 2 2
⇒
log x 2
= log x
log2 x + log x 2 − 2
= log
log x ⇒ 2
149.
x
log2 x + log x 2 − 2
= 2 log x
3 log 2 x – log28x + 1 = 0
⇒ 3 log 2 x – (log223 + log2x) + 1 = 0 ⇒ 3 log 2 x – 3 – log2x + 1 = 0 ⇒ 3 log2 x – 2 – log2x = 0 2
⇒ 3 log2 x = (2 + log2x)2
⇒9 log2x = 4 + 4 log2x + log 22 x
⇒
⇒
9 log2x – 4 log2x = 4 + log22x 0 = log22x – 5 log2x + 4
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics ⇒ 0 = log22x – log2x – 4 log2x + 4
⇒
0 = log2x (log2x – 1) – 4(log2x – 1) log2x = 4 = 16 =2 log2x = 1
∴
150.
⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
A. I. Prilepko Solution
log2x – 3 log x = log (x2) = 4 log2x – 3 log x – 2 log x + 4 = 0 log102x – 5 log10x + 4 = 0 log102x – log10x – 4 log10x + 4 = 0 (log10x – 4) (log10x – 1) = 0 log10x = 4 x=
log10x = 1
104
x = 102
151.
log1/3x – 3 log1 / 3 x + 2 = 0
⇒
– log3x – 3 − log 3 x + 2 = 0
⇒
(2 – log3x)2 = 3 − log3 x
⇒ ⇒ ⇒ ⇒ ⇒
4 + log32x – 4 log3x = – 9 log3x
⇒
x=
152.
2 (log x 5 )2 – 3 logx 5 +1 = 0
2
log32 x + 5 log3x + 4 = 0 (log3x + 4) (log3x + 1) = 0 log3x = – 4
log3x = – 1
3–4/3 =
3–1 = x
x
1 1 , 81 3
2
⇒
(log 2
⇒
x)
5
2 log52
2
–
3 +1=0 log 5 x
2 – 2 log x + 1 = 0 x 5
⇒ ⇒
0 = 2 log52x – 3 log5x + 1
⇒
log5x =
⇒
x=
153.
log22 x + 2 log2x – 2 = 0
⇒
log22 x + log2x – 2 = 0
⇒ ⇒ ⇒
(log2x – 1) (log2x + 2) = 0
0 = 2 log5x(log5x – 1) – 1(log5x – 1) 1 2
5
log2x = 1 x=
21
log5x = 1 x=5
log2x = – 2 x = 2–2
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 1 4
⇒
=2
154.
(a logb x )2 – 5 x logb x + 6 = 0
=
A. I. Prilepko Solution
Let x logba = t
⇒
( x logb a )2 – 5 x logb a + 6 = 0
⇒ ⇒⇒ ⇒
t2 – 5t + 6 = 0 t2 – 2t – 3t + 6 = 0 t(t – 2) – 3(t – 2) = 0 (t – 3) (t – 2) = 0
⇒
log a t=3= x b
⇒
t=2= log 2 = x = 2logb a
x= 1 x
155.
log2 (100x) + log2 (10x) ± 14 + log
⇒
2 2 2 2 log10 10 2 + log10 x + log10 10 + log10 x = 14 + log
⇒
4 + log102 x + 1 + log102 x = 14 + log
156.
log4 (x + 3) – log4 (x – 1)
⇒
log4
x+3 = 2 – log 22 23 x −1
⇒
log4
x+3 3 =2– x −1 2
⇒
log4
x+3 1 = x −1 2
⇒
41/2 =
⇒ ⇒ ⇒
2(x – 1) = (x + 3)
157.
log4(x2 – 1) – log4(x – 1)2 = log4 4 − x 2
1 x
1 x logbxa 3 xlog logb a
x+3 x −1
2x – 2 = x + 3 x=5
1 ( x − 1)( x + 1) (4−x2 ) 2 = log 4 ( x − 1)( x − 1)
⇒
log4
⇒
log4
⇒
2 x + 1 = 4 − x 2 x − 1
x +1 = log4 (4 – x2)1/2 x −1 2
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics x 2 + 1 + 2x x 2 + 1 − 2x
158.
2 log3
= 4−x
A. I. Prilepko Solution
2
x−3 x −3 + 1 = log3 x−7 x −1 2
x−3 x−3 + 1 = log3 log3 x − 7 x −1 2
x−3 x −3 =–1 – log3 log3 x −1 x −7
log3
( x − 3) 2 ( x − 7)
2
( x − 3)( x − 1) ( x − 7) 2
×
x −1 =–1 x −3
= 3–1
x 2 − 4x + 3 2
x − 14 x + 49
=
1 3
3x2 – 12x + 9 = x2 + 14x – 49 ⇒ 2x2 + 2x – 40 = 0 ⇒ x2 + x – 20 = 0 ⇒ x (x + 5) – 4(x + 5) = 0 (x – 4) (x + 5) = 0 x = 4, – 5 since 4 can’t satisfied the equation ∴ – 5 is a solution 159.
2 log4 (4 – x) = 4 – log2(–2 – x) 2 log22 (4 – x) + log2(–2 – x) = 4 log2 (4 – x)(–2 – x) = 4 24 = x2 – 2x – 8 0 = x2 – 2x – 8 – 16 0 = (x – 6) (x + 4) ∴ x = 6, –4
160.
3 + 2 logx + 1 3 = 2 log3(x + 1) Let log3(x + 1) = t
⇒
2 3 + log ( x + 1) = 2 log3(x + 1) 3
⇒
=3+
⇒ ⇒ ⇒ ⇒ ⇒
3t + 2 = 2t2
2 = 2t t
0 = 2t2 – 3t – 2 0 = 2t2 – 4t + t – 2 0 = 2t (t – 2) + l (t – 2) 0 = (2t + 1) (t – 2) ∴
log3(x + 1) = − x + 1 = 3–1/2
1 2
or log3x + 1 = 2 x + 1 = 32
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics x=
1
–1
3
A. I. Prilepko Solution
x=8
3 +3 2– 3 161.
logx 9x2 . log32x = 4 2 (logx9 + logxx2) log3 x = 4
(logx32 + 2) log32x = 4 (2 logx 3 + 2) (log32x) = 4 2 2 log3 x × log32x + 2 log3 x – 4 = 0
2 log3x + 2 log32 x – 4 = 0 2 log32 x + 4 log3x – 2 log3x – 4 =0 (2 log3x – 2) (log3x + 2) = 0 log3x = –2 or log3x = 1 1 9
x=3
log12/ 2
x2 (4x) + log2 8
x=
162.
=8
– log22 4x + log2x2 – log28 = 8 – (log22 4 + log22x) + 2 log2x – 3 = 8 – 2 log222 – log22x + 2 log2x – 11 = 0 163.
log 0.5 x x2 – 14 log16xx3 + 40 log4x
x =0
10
164.
6 – (1 + 4.9
4 − 2 log
4 6 – 1 + 4.9 4 log 9 39
3
3
) . log7 x = logx7
1 log7x = log x 7
4.36 6 – 1 + 2 log3 3 4 3
1 log x = 7 log7 x
36 6 – 1 + 4. 8 3
165.
log3 (4.3x – 1) = 2x + 1 32x + 1 = 4.3x – 1 32x . 31 = 4.3x – 1 Let 3x = t 1 3
= 3t2 – 4t – 1 = 0
3x =
3t2 – 3t – t + 1 = 0 (3t – 1) (t – 1) = 0
3x = 1
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics t= 166.
1 ,1 3
A. I. Prilepko Solution
x=0
log3 (3x – 6) = x – 1 3x – 1 = 3x – 6 3x = 3x – 6 3
3x = 9 x = 22
t =t–6 3
t = 3t – 18 – 2t = – 18 t=9 167.
168.
log3 (4x – 3) + log3 (4x – 1) = 1 Let 4x = t log3(4x – 3) (4x – 1) = 1 31 = (t – 3) (t – 1) 3 = t2 – 4t + 3 0 = t (t – 4) t = 0, 4
Not possible 4x = 4 x=1
log3 (log1/22x – 3 log1/2x + 5) = 2 log 22 −1 x – 3 log2 −1 x + 5 = 32
⇒
∴
– log22x + 3 log2x = 4 0 = log22x + 3 log2x + 4 0 = (log2x – 1) (log2x + 4) log2x = 4 log2x = 4 x=2 x = 16 2
169.
log5
2+ x = log 5x +1 10
log5
2+x 2 – log5 =0 10 x +1
2 x +1 × x +1 2 log5
=0
2x + x 2 + 2 + x = 50 20
x2 + 3x + 2 = 20 x2 + 3x – 18 = 0 + 6x – 3x (x + 6) (x – 3) = 0 x = 3, – 6 Not satisfied 3 Ans. 170.
1 + 2 logx + 25 = log5 x + 2 Let log5x + 2 = t
5–1 = x + 2 x=
1 9 –2=– 5 5
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 2 1 + log x + 2 = log5 x + 2 5
1+
2 =t t
172.
log5x + 2 = 2 52 = x + 2
t + 2 = t2 0 = t2 – t – 2 0 = (t – 2)–2 +1 (t + 1) t = –1, 2 171.
x = 23
log424x = 2log2 4 log424x = 4 24x = 44 44x = (22)4 24x = 8 4x = 8 x=2
log2
15 = log x − 1 2 8
Let log2x = t
⇒
15 log2x – log24 = log x − log 8 − 1 2 2
⇒
15 log2x – 2 = log x − 3 − 1 2 x 4
15 log2x – 2 = log x − 4 2
15 ( t − 4)
(t – 2) =
(t – 2) (t – 4) = 15 t2 – 2t – 4t + 8 = 15 t2 – 6t + 8 – 15 = 0 t2 – 6t – 7 = 0 t2 – 7t + t – 7 = 0 t (t – 7) + l(t – 7) = 0 (t + 1) (t – 7) = 0 t = –1, 7 log2x = –1 or x=
173.
A. I. Prilepko Solution
1 2
1 − 2(log x 2 )2 log x − 2(log x )2
log2x = 7 x = 27
=1
2 1 − 2( 4 log10 x) 2 log10 x − 2 log10 x
=1
1 – 8 log102x = log10x – 2 log102x
Let log x = t
log10x = 0 x = 10°
0 = 6log10 2 x + log10x – 1 Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 0 = (3 log10x – 1) (2 log10x + 1) = 0 ∴ 3 log10x = 1 or 2 log10x = – 1 log10x = x= 174.
3
1 3
log10x = – x=
10
(4.3x
(9x
log2 – 6) – log2 – 6) = 1 log2(4.t – 6) – log2 (x2 – 6) = 1 log2
4t − 6 t2 − 6
A. I. Prilepko Solution
1 2
1 10
Let 3x = t
=1
4t – 6 = 2(t2 – 6) 0 = 2t2 – 12 – 4t + 6 0 = 2t2 – 4t – 6 0 = t2 – 2t – 3 t2 – 3t + t – 3 = 0 ∴ t (t – 3) + 1(t – 3) = 0 (t + 1) (t – 3) = 0 ∴ t = 3, – 1 3x = 3 x=1 3x = – 1 Rejected ∴ Answer = 1
175.
1 log (5x – 4) + log 2
0.18 x + 1 = 2 + log 100
1 1 log10 (5x – 4) + log10x + 1 = 2 + log1018 – log10100 2 2 1 log10 (5x – 4) (x – 1) = log1018 2 5x2 – 4x + 5x – 4 = 324 5x2 + x – 328 = 0 2 D = b – 4ac = 1 + 6560 = 6561
⇒
D = 6561
x≡– ∴ 176.
82 ,8 10
Final Ans. 8
log4 (2.4x – 2 – 1) + 4 = 2x
21x log22 2. 2 − 1 + 4 = 2x 4 2.2 2x − 16 1 log2 + 4 = 2x 16 2
2 2x +1 − 16 log2 + 4 = 4x 16
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
2 2 x +1 − 16 16
177.
logx 5 + logx 5x – 2.25 = (logx 5 )2 logx
11 5 + logx5 + logxx – 2.25 = 2 log x 50
2
1 1 1 + + 1 – 2.25 = 2 log5 x log5 x 4 log5 x
Let log5x = t
1 1 1 + + 1 – 2.25 = 2t t 4t
=
1 + 2 + 2t − 2t(2.25 ) 1 = 2t 4t
2 + 4 + 4t – 4t (2.25) = 1 6 + 4 (t – 2.25 t) = 1 – 6 –1 – 1.75 t = 178.
−5 5 = 4 7
log (log x) + log (log x4 – 3) = 0 log10 [log x + 4 logx – 3] = 0 5 log x – 3 = 1 5 log10x = 4 log10x =
4 5
Let logx = t
x = 104/5 ⇒ 5 log2x – 3 log ? = 0 log10t + 4 log2t – 4 log 3 179.
180.
log3x – 2 log1/3 x = 6 log3x + 2log3x = 6 log3x + log3x2 = 6 x3 = 36 x3 = (32)3 x=9 2 log x log(5 x − 4) = 1
⇒
2 log x = log (5x – 4) log x2 = log (5x – 4) x2 – 5x + 4 = 0 x2 – 4x – x + 4 = 0 x(x – 4) –1(x – 4) = 0 (x – 1) (x – 2) = 0 x = 1, 4 Answer = 4
181.
2 log82x + log8x2 + 1 – 2x = log8(2x)2 + log8 x2 – 2x + 1 =
4 3
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Mathematics Online: It’s all about Mathematics log84x2 (x2 – 2x + 1) =
A. I. Prilepko Solution
4 3
4x2(x2 – 2x + 1) = (23) 4x2(x2 – 2x + 1) = 24 4x2 (x2 – 2x + 1) = 16 x2 [x2 + 2x + 1] = 4 x4 + 2x3 + x2 – 4 = 0 2 log82x + log8(x2 + 1 – 2x) =
4 3
2[log82 + log8x] + log8 x2 – 2x + 1 =
4 3
(x + 1) (x3 – 3x2 + 4x – 4) (x – 2) (x2 – x + 2) Thus (x – 1) (x – 2) (x2 – x + 2) are its factor since 1 can’t satisfies, 2 satisfies the answer ∴ Answer = 2
182.
1 1 log2(x – 2) – = log1/8 6 3
3x − 5
1 1 log2 (x – 2) = = log2−3 6 3
3x − 5
1 1 1 log2 (x – 2) – =– log2(3x – 5)1/2 6 3 3 1 1 log2 (x – 2)1/6 . (3x – 5)1/6 = 6 3
= log2 (x – 2)1/6 . (3x – 5)1/6 =
1 3
((x – 2)(3x – 5))1/6 = (2)1/3 × 6 if 9 multiplied power by, get 3x2 – 6x –5x + 10 = 4 3x2 – 11x + 6 = 0 3x (x – 3) – 2(x – 3) = 0 x=
2 ,3 3
Since ∴ 183.
2 doesn’t satisfic equation 3
final answer = 3
2 log3(x – 2) + log3 (x – 4)2 = 0 log3(x – 2)2 + log3 (x – 4)2 = 0 2 log3(x – 2) + 2 log3 (x – 4) = 0 2[log3(x – 2) (x – 4)] = 0 x2 – 2x – 4x + 8 = 3° x2 – 6x + 8 – 1 = 0 x2 – 6x + 7 = 0
D = b2 – 4ax 36 – 28 = 8 D =2 2
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Mathematics Online: It’s all about Mathematics ∴
x=
6±2 2 2
=2
(3 ± 2 ) 2
=3+ 2,3– 184.
A. I. Prilepko Solution
2
log2 (2 x 2 ). log 4 16 x = log4x3
let log2x = t
(log2 2 + log 2 x 2 )(log4 16 + log 4 x ) = 3 log4x
⇒
1 (1 + 2 log 2 x ) 2 + log2 x = 3 log x 2 2 2
(1 + 2t ) 2 + t = 2
2
3 = t 2
2
2 4+t = 9t (1 + 2t) 2 4
2 (4 + 8t + t + 2t2) = 9t2 0 = 9t2 – 4t2 – 18t – 8 0 = 5t2 – 18t – 8 0 = 5t2 – 20t + 2t – 8 0 = 5t (t – 4) + 2(t – 4) 0 = (5t + 2) (t – 4) t = 4, –
∴
log2x = 4
x = 24 x = 16
185.
186.
2 5
∴
log2x = –
x = (2)–2/5
2 5
can’t satisfied me
3 log x + 19 3 log x − 1 = 2 logx + 1
Let log10x = t
3 t + 19 = 2t + 1 3t − 1
log10x = 2
3t + 19 = (2t + 1)(3t – 1)
log10x = –
3t + 19 = 6t2 + 3t – 2t –1 0 = 6t2 – 2t – 20 0 = 3t2 – t – 10 0 = 3t2 – 6t + 5t – 10 0 = (3t + 5) (9t – 2)
x = 10– 5/3
log( x + 1 + 1) log 3 x − 40
5 3
=3
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Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
3 log ( x + 1 + 1) = 3 log x − 40 x + 1 + 1 = logx – 120
log
120 = logx – log
x +1 – 1
187.
log326 – log32 2 = (log2x – 2) log3121 log32 2 + log323 – log322 = (log2x – 2) log312 (log3x)2 = (log102x – 2) (log33 + log34)
188.
1–
1 1 log (2x – 1) = log (x – 9) 2 2
1=
1 1 log (x – 9) + log (2x – 1) 2 2
1 [log (x – 9) (2x – 1)] 2 2 = log10 2x2 – 19x + 9 102 = 2x2 – 19x + 9 0 = 2x2 – 19x – 91 0 = 2x2 – 26x + 7x – 91 0 = 2x (x – 13) + 7(x – 13) 0 = (2x + 7) (x – 13)
1=
7 , 13 2 x = 13 Ans.
x=–
189.
log x + 7 − log 2 =–1 log 8 − log( x − 5)
log10
x+7 2 8
log10 x −5
log10 190.
=–1
8 x+7 = ? – log10 x −5 2
log(3x2 + 7) = log (3x – 2) = 1 log (3x2 + 7) = 1 log (3x – 2) = 1 101 = 3x2 + 7 101 = 3x – 2 3 = 3x2 12 = 3x x=1 x=4 log (3x2 + 7) – log (3x – 2) = 0 log
3x 2 + 7 = 0 3x − 2
3x2 + 7 = 3x – 2 3x2 – 3x + 9 = 0 x2 – x + 3 = 0
191.
1 log(x2 – 16x + 20) – log 3
3
1 7 = 3 log (8 – x)
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Mathematics Online: It’s all about Mathematics
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1 1 [log10x2 – 16x + 20 – log107] = log108 – x 3 3
log10
x 2 − 16x + 20 = log108 – x 7
x2 – 16x + 20 = 56 – 7x x2 – 16x + 20 – 56 + 7x = 0 x2 – 9x – 36 = 0 – 12 + 3 x(x – 12) + 3(x – 12) = 0 (x + 3) (x – 12) = 0 ∴ x = –3, 12 makes the x < 1 rejected x = –3 Ans. 192.
log63x + 3 – log6 3x – 2 = x log6 2x + 3 3x − 2 2x + 3 3x − 2
2x + 3 3x − 2
=x
= 6x
= 2x . 3x
2x 3x 9 2x × x = 2x .3x 3x = 3 8 32
193.
1 1 + log 3 + log 2 = log (27 – 2x
194.
log(5x – 2 + 1) = x + log 13 – 2 log 5 + (1 – x) log 2
196.
log2(4x + 1) = x log2(2x + 3 – 6)
2
3)
2x log2 (22x + 1) = x + log2 8 − 6 198.
log3 (9x + 9) = x + log3 (28 – 2 . 3x)
199.
log (log x) + log (log x3 – 2) = 0 log10 [log x + logx3 – 2] = 0 4 log x – 2 = 1 log x =
Pattern may be same of Q.No. 178
3 4
x = 103/4 Answer has been given is 110.
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A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 200.
log log
5
4x − 6 5
2 log5
log5
4x – 6 = – log
2x − 2
5
A. I. Prilepko Solution
2x– 1 = 2 Let 2x = t
=2
t2 − 6 =2 t−2
2x = 4
t2 − 6 =1 t−2
2x = 4
t2 − 6 = 5a1 t−2
∴
x=6
x=2
t2 – 6 = 5t – 10 t2 – 5t + 4 = 0 t(t – 1) – 4(t – 1) = 0 t = 4, 1
203.
x 24 log 3 − x 2
204.
1 5
205.
x
206.
log2 (25x + 3 – 1) = 2 + log2 (5 x + 3 + 1)
210.
log2(2x2) . log2 (16x) =
1 log 4 x = 2 + 4 log 16 – 2
log2 x −log x
=
2 3 log2 x − log x 3
1 . 5logx – 1 125
3 = 100 10
9 log22x 2
(log22 + log2x2) (4 + log24) =
9 log22 x 2
9 log22x 2 2 [4 + 8 log2x + log2x + 2log22x] = 9 log22x 8 + 16 log2 x + 2 log2x + 4 log22x = 9 log22 x 0 = 5 log22x – 18 log2x – 8 Let log2x = t 0 = 5t2 – 18t – 8t D = b2 – 4ac = 324 + 160 = 484
(1 + 2 log2x) (4 + log2x) = ⇒ ⇒
D = 22
t= ∴
18 ± 22 40 = =4 10 10
log2x = 4
log2x = –
x = 24
x = (2)– 2/5
2 5
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Mathematics Online: It’s all about Mathematics 16 = 2 – 2/5
211.
1 log3 = log 3 3 + 27 log44 + 1 + 2 x since radical should be defined 1+
1 24
= log x 3 + 27 log (4.3 . 31/2x) = log (31/x + 27) 12.t = t2 + 27 0 = t2 – 12t + 27 0 = t2 – 9t – 3t + 27 0 = (t – 9) (t – 3) ∴ t = 9, 3 log44 + log 3
Let 31/x = t 31/x = t
1 2
x = 1,
212.
A. I. Prilepko Solution
log(x3 + 27) – log (x2 + 6x + 9) = log log (x + 3) (x2 – 3x + 9) – log
3
7
( x 2 + 6x + 9)1/ 2 ( x + 3) 2
= log 7
( x + 3)( x 2 − 3 x + 9) = log 7 10 ( x + 3)
log10
x2 – 3x + 9 – 7 = 0 x2 – 3x + 2 = 0 x2 – 2x – x + 2 = 0 x(x – 2) – 1(x – 2) = 0 (x – 1) (x – 2) = 0 x = 1, 2 213.
215.
5log x = 50 – xlog 5 5log x + xlog 5 = 50 5log x + 5log x = 50 5t + 5t = 50 2.5t = 50 5t = 25 t=2
x
3 log x −
1 log x
=
3
Let log10x = t x = 102 = 100
10
2
x
3 log x −1 log x
log x
=
3
3 log2 x −1 log x
log x .
10 = log
3
10
3 log2 x − 1 1 = log 10 log x 3
3 log2x – 1 =
1 3
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A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 3 log2x =
4 3
log102x =
4 9
log10x = ±
2 3
A. I. Prilepko Solution
x = (10)2/3 , (10–2/3 216.
|x – 10| log2 (x – 3) = 2(x – 10) log2 (x – 3)|x – 10| = 2(x – 10) (x + 10) log2(x – 3) = 2(x – 10)1 – (x + 10) log2(x – 3) = 2(x – 10) log2(x – 3) = – 2 log2(x – 3) = 2 x – 3 = 29 (x – 3) = 2–2 x=7
217.
(x – 3) =
1 4
⇒x=
1 13 +3 = 4 4
log4 log2x + log2 log4x = 2
log2x = t
1 1 log2 log2x + log2 log2x = 2 2 2
2x = 4
1 t log2t + log 2 =2 2 2 log2 t1/2 + log2t – log22 = 2 log2t1/2 + log2t – log22 = 2 log2t1/2 + log2t = 3
x = 24
1 log2t + log2t = 3 2 log2t + 2 log2t = 6 3 log2 t = 6 t = 22 = 4
218.
(6x – 5) |ln (2x + 2.3)| = 8 ln (2x + 2.3) 8ln (2 x + 2.3) (6x – 5) = | ln (2x + 2.3) |
Ist
6x – 5 = – 8 6x = 13 x=
IInd
13 1 =– 6 2
It is not in answer
6x – 5 = 8
The right answer
6x = 13
=
x=
13 −13 , 6 20
13 6
log9 9 x 8 − log3 3 x = log3x3
219. =
log9 9 + log9 x 8 .(log3 3 + log3 x ) = 3 log3x
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A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics
A. I. Prilepko Solution
2
8 = 1 + log3 x.(1 + log3 x ) = (3 log3x)2 2 = (1 + 4 log3x) (1 + log3x) = 9 log32 x (1 + 4t) (1 + t) = 9t2 0 = 9t2 – 4t2 – 5t – 1 0 = 5t2 – 5t – 1 D = b2 – 4ac = 25 + 20 = 45 5±3 5 10
t=
log3x = t log3 x =
5±3 5 10
x=3
220.
5±3 5 10
log2 (100x) – log2 10x + log2x = 6 2 2 log102 102 + log102x – [log10 10 + log10 x] + log102x = 6 2 2 − 1 − log10 x =6 4 + log10 2 x= 3 log10
log10x = 221.
3
9 log1/ 3 ( x +1) = 5log1/ 5 ( 2 x
3
2 log
3 −1
( x +1)
= 5
log
5−1
2
+1)
( 2 x 2 +1)
x−7 x −1
−2
2 –1 3 −2 log3 ( x +1) = (2x + 1) (x + 1)–2 = (2x2 + 1)–1
1 ( x + 1)2
2x2
=
1 2
2x + 1
x2
+1= + 1 + 2x 2x2 – x2 – 2x = 0 x2 – 2x = 0 x(x – 2) = 0 x = 0, 2
223.
2 log2
log2 ⇒ ⇒ ⇒
x − 1 =1 + log2 x + 1
x 2 + 49 − 14 x x −1 × =1 ( x − 1)( x − 1) x +1
x 2 + 490 − 14 x x2 − 1
=2
x2 + 49 – 14x = 2x2 – 2 x2 + 14x – 2 – 49 = 0 x2 + 14x – 51 = 0 x2 + 17x – 3x – 51 = 0
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Mathematics Online: It’s all about Mathematics x(x + 17) – 3(x + 17) = 0 ∴ x = – 17, 3→can’t satisfy the equation. ∴ x = – 17 only
225.
A. I. Prilepko Solution
1 3 1 log (1 + x) + log (1 – x) = log (1 – x2) 2 2 2
=
1 1 [log (1 + x) + log (1 – x)3] = log (1 – x2) 2 2
log
⇒
1+ x (1 − x )2 (1 − x )
log
= log (1 – x2)
1+ x (1 − x )2 (1 − x ) 1+ x
log10
(1 − x )3 1− x2
(1 + x ) (1 − x )
log10
×
3
– log(1 – x2) = 0
=0
1 =1 (1 + x )(1 − x )
1 (1 − x )4
=0
log10 (1 – x)–4 = 0 – 4 log10 (1 – x) = 0 100 = 1 – x x=0
226.
log(35 − x 3 ) =3 log(5 − x ) log (35 – x3) = 3 log (5 – x) log (35 – x3) = log(5 – x)3 35 – x3 = 125 – x3 – 15x (5 – x) 35 – 125 = – 75x + 15x2 0 = 15x2 – 75 x + 90 0 = x2 – 5x + 6 0 = x2 – 2x – 3x + 6 ∴ x = 2, 3
227.
logx2 – log4x +
7 =0 6
1 7 1 log2 x – 2 log2x + 6 = 0
6 – 3 log22 x ± 7 log2x = 0 3 log22 x – 9 log2x + 2 log2x – 6 = 0 (3 log2x + 2) (log2x – 3) = 0 3 log2x = – 2 log2x = 3 log2x = – x = 2 – 2/3
2 3
x = 23 x=8
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Mathematics Online: It’s all about Mathematics 229.
A. I. Prilepko Solution
log 2 x – 0.5 = log2 x log 2 x =
1 1 log2x + 2 2
1 [log2x + 1] 2 4 log2x = log22x + 1 + 2 log2x 0 = log22 x – 2 log2x + 1 0 = log2x (log2x – 1) – 1(log2x – 1) 0 = log2x – 1 1 = log2x x=2 log 2 x =
230.
log1/3
1 x 2 × − 1 = log 1/3 2 x
1 1 2 –1= 2 2
1 2x − 4 2
2x
x
x
–4
Let
1 =t 2
x
1 1 1 2 – . =–3 2 2 2
⇒ ⇒ ⇒ ⇒ ⇒
2t – t2 = 3 0 = t2 – 2t – 3 0 = t(t + 1) – 3(t + 1) (t – 3) (t + 1) = 0 t = 3, –1
Q
1 =t 2
x
x
1 = – 1, 3 2
–1 not defined
x
1 log = log 3 2
log 2–x = log 3 – x log 2 = log 3 log 3 – x = log 2
x = – log23 Ans.
231.
1 1 log6(x – 2) + log6(x – 11) = 1 2 2 1 log6 (x – 2) (x – 11) = 1 2 log6 x2 – 13x + 22 = 2 x2 – 13x + 22 = 62 x2 – 13x + 22 – 36 = 0 x2 – 13x – 14 = 0 x2 – 14x + x – 14 = 0 x(x – 14) + 1(x – 14) = 0
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Mathematics Online: It’s all about Mathematics (x + 1)(x – 14) = 0 x = 14 Ans. –1 is not defined 233.
A. I. Prilepko Solution
1/ 2 log7 2 + log72 x = log 7 −1 (3)
log72 +
1 1 log7x = – log73 2 2
1 1 log73 + log7x = 0 2 2 2 log72 + log7 3 + log7x = 0 log7 12 = – log7x x = – 12
log72 +
237.
log3x
3 + log32x = 1 x
3 x + log 2 x = 1 3 log3 3 x log3
Let log3x = t
log3x = 0
log3 3 − log3 x 2 log3 3 + log3 x + log 3 x = 1
log3x = 1
1− t + t2 = 1 1+ t
x=1
1 – t + t2 (1 + t) = 1 + t t3 + t2 – 2t = 0 t(t2 + t – 2) = 0 t2 + t – 2 = 0 t(t + 2) – 1(t + 2) = 0 t = 1, 2 Ans. 1, 3, 3–2 246.
x = 3–2
log4(x + 12) logx2 = 1 x + 12 1 log log x = 1 2 2
log2 (x + 12) = 2 log2x log2 x + 12 = log2x2 x + 12 = x2 0 = x2 – x – 12 0 = x2 – 4x + 3x – 12 0 = x(x – 4) + 3(x – 4) x = 4, 3 not satisfied ∴ 24 Ans. 247.
log16x + log4x + log2x = 7 1 1 log2x + log2x + log2x = 7 4 2 log2 x + 2 log x + 4 log2 x =7 4 7 log2x = 28 log2x = 4
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Mathematics Online: It’s all about Mathematics x = 16
248.
log 10 +
1 log (3 2 3
1 2 log (3 3
log10 (3 2
x
x
x
+ 271) = 2
+ 271) = 1
+ 271) = 3
32
x
+ 271 = 103
32
x
= 1000 – 271
32
x
= 279
32
x
= 36
x =3
⇒
2 x =6
⇒
( x )2 = 9
250.
log5 (3x + 10) + 7.10 = log5 (9x + 156) Detective Question for me and how can I solve
251.
(log4x – 2) log4x =
3 (log4x – 1) 2 2(log4x – 2) log4x = 3 log4x – 3 (2t – 4) t = 3t – 3 2t2 – 4t – 3t + 3 = 0 2t2 – 7t + 3 = 0 2t2 – 6t + t + 3 = 0 2t (t – 3) – 1(t – 3) = 0
t = 3,
A. I. Prilepko Solution
⇒
x=9
Let log4x = t
1 2
∴
log4x = 3,
1 2
x = 43 , (4)1/2 x = 64, 2 252.
log5x + log25x = log1/5
3
1 1 log5x = – log5 3 2 2 2 log5x + log5x = – log 53 – 3 log5x = log53 log5x–3 = log5 3 (x–3)1/3 = (3)– 1/3
log5x +
x= 254.
1 3
3
log2 (9 – 2x) = 3 – x 23 – x = 9 – 2x
Let 2x = t
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A. I. Prilepko Solution
Mathematics Online: It’s all about Mathematics 23 2x
= 9 – 2x
2x = 1
8 =9–t t
2x = 8
8 = 9t – t2 t2 – 9t + 8 = 0 t2 – 8t – t + 8 = 0 t (t – 8) – 1 (t – 8) = 0 (t – 1) (t – 8) = 0 t = 1, 8 x = 0, x = 3 Ans. 255.
256.
2(log 2 – 1) + log (5
x
2(log 2 – 1) + log (5
x
logx3 + log3x = log
A. I. Prilepko Solution
x
x=3
+ 1) + log (51−
x
+ 5)
5 + 5 + 1) + log 5 x
3 + log3 x +
1 2
1 2 1 1 log3 x + log3x = log3 x + 2 log3x + 2 1 2 t 1 +t= + + t t 2 2
1+ t 2 2 + t2 + t = t 2t
2 + 2t2 = 2 + t2 + t t2 – t = 0 t = 0, 1
259.
Q
log3x = 0 x = 30 = 1 log3x = 1
logx 125x . log225 x = 1
(log5 2 x )2
log5 125 x 2 log5 x . log5 2x = 1
1 log5 x 2
⇒
x=3
2
log5 53 + log5 x 1 . log25x = 1 log5 x 2 3+t t2 . =1 t 4
3t 2 + t 3 =1 4t
t(3t + t2) = 4t t2 + 3t – 4 = 0 t2 + 4t – t – 4 = 0 t(t + 4) – 1(t + 4) = 0 (t – 1) (t + 4) = 0 t = 1, – 4 ∴ log5x = 1, –4 = 5, 5–4It’s all about Mathematics MathematicsxOnline:
A. I. Prilepko Solution